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Logarithm Questions

Logarithm questions with answers are provided for students to solve them and understand the concept elaborately. These questions are based on the logarithm chapter of Class 9, 10 and 11 syllabi. Practising these problems will not only help students to score good marks in academic exams but also participate in competitive exams conducted at the state or national level, such as Maths Olympiad.

The logarithmic function is an inverse of the exponential function. It is defined as:

y=log a x, if and only if x=a y ; for x>0, a>0, and a≠1.

Natural logarithmic function: The log function with base e is called natural logarithmic function and is denoted by log e .

f(x) = log e x

The questions of logarithm could be solved based on the properties, given below:

Also, read:

  • Logarithm Table

Questions on Logarithm with Solutions

1. Express 5 3 = 125 in logarithm form.

As we know,

a b = c ⇒ log a c=b

Log 5 125 = 3

2. Express log 10 1 = 0 in exponential form.

Given, log 10 1 = 0

By the rule, we know;

log a c=b ⇒ a b = c

3. Find the log of 32 to the base 4.

Solution: log 4 32 = x

(2 2 ) x = 2x2x2x2x2

log 4 32 =5/2

4. Find x if log 5 (x-7)=1.

Solution: Given,

log 5 (x-7)=1

Using logarithm rules, we can write;

5. If log a m=n, express a n-1 in terms of a and m.

6. Solve for x if log(x-1)+log(x+1)=log 2 1

Solution: log(x-1)+log(x+1)=log 2 1

log(x-1)+log(x+1)=0

log[(x-1)(x+1)]=0

Since, log 1 = 0

(x-1)(x+1) = 1

Since, log of negative number is not defined.

Therefore, x=√2

7. Express log(75/16)-2log(5/9)+log(32/243) in terms of log 2 and log 3.

Solution: log(75/16)-2log(5/9)+log(32/243)

Since, nlog a m=log a m n

⇒log(75/16)-log(5/9) 2 +log(32/243)

⇒log(75/16)-log(25/81)+log(32/243)

Since, log a m-log a n=log a (m/n)

⇒log[(75/16)÷(25/81)]+log(32/243)

⇒log[(75/16)×(81/25)]+log(32/243)

⇒log(243/16)+log(32/243)

Since, log a m+log a n=log a mn

⇒log(32/16)

8. Express 2logx+3logy=log a in logarithm free form.

Solution: 2logx+3logy=log a

Video Lesson

Logarithmic equations.

logarithm problems and solutions

9. Prove that: 2log(15/18)-log(25/162)+log(4/9)=log2

Solution: 2log(15/18)-log(25/162)+log(4/9)=log2

Taking L.H.S.:

⇒2log(15/18)-log(25/162)+log(4/9)

⇒log(15/18) 2 -log(25/162)+log(4/9)

⇒log(225/324)-log(25/162)+log(4/9)

⇒log[(225/324)(4/9)]-log(25/162)

⇒log[(225/324)(4/9)]/(25/162)

⇒log(72/36)

⇒log2 (R.H.S)

10. Express log 10 (2) + 1 in the form of log 10 x.

Solution: log 10 (2)+1

11. Find the value of x, if log 10 (x-10)=1.

Solution: Given, log 10 (x-10)=1.

log 10 (x-10) = log 10 10

12. Find the value of x, if log(x+5)+log(x-5)=4log2+2log3

log(x+5)+log(x-5)=4log2+2log3

log(x 2 -25) = log2 4 +log3 2

log(x 2 -25) = log16+log9

log(x 2 -25)=log(16×9)

log(x 2 -25)=log144

x 2 -25=144

13. Solve for x, if (log 225/log15) = log x

Solution: log x = (log 225/log15)

log x = log 15 2 /log 15

log x = 2log 15/log 15

x=10×10

Practice Questions

  • If log x = m+n and log y=m-n, express the value of log 10x/y 2 in terms of m and n.
  • Express 3 -2 =1/9 in logarithmic form.
  • Express log 10 0.01=-2 in exponential form.
  • Find the logarithm of 1/81 to the base 27.
  • Find x if log 7 (2x 2 -1)=2.

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LOGARITHM PROBLEMS WITH SOLUTIONS

Problem 1 :

Find the logarithm of 64 to the base 2√2.

Write 64 as in terms of 2√2.

= 2 4  ⋅  2 2

= 2 4  ⋅  [(√2) 2 ] 2

= 2 4  ⋅ (√2) 4

log 2√2 64 =  log 2√2 (2√2) 4

= 4 log 2√2 (2√2)

Problem 2 :

If log a bc = x, log b ca = y and log c ab = z, then find the value of

x + 1 = log a bc + log a a = log a abc

y + 1 = log b ca + log b c = log b abc

z + 1 = log c ab + log c c = log c abc

Problem 3 :

If a = log 24 12, b = log 36 24 and c = log 48 36, then find the value of (1 + abc) in terms of b and c.

1 + abc = 1 + log 24 12  ⋅  log 36 24  ⋅  log 48 36

= 1 + log 36 12  ⋅  log 48 36

= 1 + log 48 12

= log 48 48 + log 48 12

= log 48 (48  ⋅  12)

= log 48 (2  ⋅  12) 2

= 2log 48 24

= 2log 36 24  ⋅  log 48 36

Problem 4 :

Find the value of log 0.0001 to the base 0.1.

log 0.1 0.0001 = log 0.1 (0.1) 4

= 4log 0.1 0.1

Problem 5 :

If 2logx = 4log3, then find the value of x.

2logx = 4log3

Divide each side by 2.

logx = 2log3

logx = log3 2

logx = log9

Problem 6 :

Find the value of log √2 64.

log √2 64 = log √2 (2) 6

= 6log √2 (2)

= 6log √2 (√2) 2

= 6  ⋅  2log √2 (√2)

= 12  ⋅ 2(1)

Problem 7 :

Find the value of log ( ¹⁄₈₁ ) to the base 9.

= log 9 ( ¹⁄₈₁ )

= log 9 1 - log 9 81

= 0 - log 9 (9) 2

= -2log 9 9

Problem 8 :

Find the value of log(0.0625) to the base 2.

= log 2 (0.0625)

= log 2 (0.5) 4

= 4log 2 (0.5)

= 4log 2 ( ½ )

= 4(log 2 1 - log 2 2)

Problem 9 :

Given log2 = 0.3010 and log3 = 0.4771, find the value of log6.

= log(2  ⋅  3)

= log2 + log3

Substitute the values of log2 and log3.

= 0.3010 + 0.4771

Problem 10 :

Find the value of log(0.3) to the base 9.

= log 9 (0.3)

= log 9 ( ⅓ )

= log 9 1 - log 9 3

= 0 - log 9 3

Problem 11 :

Solve for x :

log 2 x -  log 2 5 = 2 +  log 2 3

Use the fundamental law of logarithm.

log 2 ( ˣ⁄₅ ) = 2 +  log 2 3

Subtract  log 2 3 from each side.

log 2 ( ˣ⁄₅ ) -  log 2 3 = 2

log 2 ( ˣ⁄₁₅ ) = 2

Convert to exponential form.

ˣ⁄₁₅  = 2 2

ˣ⁄₁₅  = 4

Multiply each side by 15.

Problem 12 :

Find integer values of m and n for which 

m - n log 3 2 = 10 log 9 6

Use the property of logarithm.

m - n log 3 2 = 5 log 3 6

m - n log 3 2 = 5 log 3 (3 ⋅ 2)

m - n log 3 2 = 5[ log 3 3 +  log 3 2]

m - n log 3 2 = 5[1 +  log 3 2]

Use the distributive property.

m - n log 3 2 = 5 + 5 log 3 2

Equate the constant terms and coefficients of like terms.

Problem 13 :

Solve the simultaneous equations :

log 2 6x = 1 + 2 log 2 y

1 + log 6 x =  log 6 (15y - 25)

log 2 6x = 1 + 2 log 2 y ----(1)

1 + log 6 x =  log 6 (15y - 25) ----(2)

log 2 6x =  log 2 2 + 2 log 2 y

log 2 6x =  log 2 2 + log 2 y 2

log 2 6x =  log 2 (2y 2 )

6x = 2y 2   ----(3)

log 6 6 + log 6 x =  log 6 (15y - 25) 

log 6 (6x)  =  log 6 (15y - 25) 

6x = 15y - 25

From (3), substitute 2y 2 for 6x in (4).

2y 2  = 15y - 25

2y 2  - 15y + 25 = 0

Factor and solve.

2y 2  - 10y - 5y + 25 = 0

2y(y - 5) - 5(y - 5) = 0

(y - 5)(2y - 5) = 0

y = 5  or  y =  ⁵⁄₂

Substitute 5 and ⁵⁄₂  for y in (3).

Therefore, 

x = ²⁵⁄₃ , y = 5

x = ²⁵⁄₁₂ , y =  ⁵⁄₂

Problem 14 :

log 2 (x + 3) +  log 2 (x - 3) = 4

log 2 [(x + 3)(x - 3)] = 4

(x + 3)(x - 3) = 2 4

x 2  - 3 2  = 16

x 2  - 9 = 16

x 2  = 25

√x 2  = √25

x = ± 5

x = -5 or 5

But, x = -5 will not work with the given equation.

Because, x = -5 makes the arguments negative in the logarithms.  Since logarithm is defined only for positive value in argument, x = -5 is not a solution.

So, x = 5 is the solution to the above equation.

Problem 15 :

Prove that :

We can take the expression in (1) and prove, that is equivalent to the expression in (2).

So,  (1) = (2). 

Hence, proved.

Video Lessons

Introduction to Logarithms

Fundamental Laws of Logarithms

Change of Base

Using Log Table

Using Antilog Table

Important Stuff

Difference between Bar Value and Negative Value

Multiplication of Two Logarithms

Relationship between Exponents and Logarithms

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4.6e: Exercises - Exponential and Logarithmic Equations

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A: Concepts

Exercise \(\PageIndex{1}\) 

1) How can an exponential equation be solved?

2) When does an extraneous solution occur? How can an extraneous solution be recognized?

3) When can the one-to-one property of logarithms be used to solve an equation? When can it not be used?

Determine first if the equation can be rewritten so that each side uses the same base. If so, the exponents can be set equal to each other. If the equation cannot be rewritten so that each side uses the same base, then apply the logarithm to each side and use properties of logarithms to solve.

The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.

B: Solve Exponential Equations Using the 1-1 Property (like Bases)

Exercise \(\PageIndex{2}\) 

\( \bigstar \)  For the following exercises, use like bases to solve the exponential equation.

C: Solve exponential equations using logarithms

Exercise \(\PageIndex{3}\) 

\( \bigstar \)  For the following exercises, use logarithms to solve.

\( \bigstar \)  Solve. Give the exact answer and the approximate answer rounded to the nearest thousandth .

\( \bigstar \)  Find the \(x\)- and \(y\)-intercepts of the given function.

\( \bigstar \)  Use a \(u\)-substitution to solve the following.

D: Mixed exponential equations

Exercise \(\PageIndex{4}\) 

\( \bigstar \) S olve each exponential equation. Find the exact answer and then approximate it to three decimal places.

\( \bigstar \)  For the following exercises, solve the exponential equation exactly. 

\( \bigstar \)  For the following exercises, solve each equation. Write the exact solution, and then approximate the answer to \(3\) decimal places.

E: Solve log equations by rewriting in exponential form

Exercise \(\PageIndex{5}\) 

\( \bigstar \)  For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation.

\( \bigstar \)  For the following exercises, use the definition of a logarithm to solve the equation.

\( \bigstar \)  Solve.

F: Solve log equations using the 1-1 property

Exercise \(\PageIndex{6}\) 

\( \bigstar \)  For the following exercises, use the one-to-one property of logarithms to solve.

G: Mixed log equations

Exercise \(\PageIndex{7}\) 

\( \bigstar \) S olve for \(x\). Give exact answer (not a decimal approximation).

\( \bigstar \) Solve for \(x\).

\( \bigstar \)  Solve for \(x\).

H: Inverses of Log and Exponent Functions

Exercise \(\PageIndex{8}\) 

\( \bigstar \)  Find the inverse of the following functions.

I: Mixed log and exponential equations

Exercise \(\PageIndex{9}\) 

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  • \log _2(x+1)=\log _3(27)
  • \ln (x+2)-\ln (x+1)=1
  • \ln (x)+\ln (x-1)=\ln (3x+12)
  • 4+\log _3(7x)=10
  • \ln (10)-\ln (7-x)=\ln (x)
  • \log _2(x^2-6x)=3+\log _2(1-x)
  • How do you calculate logarithmic equations?
  • To solve a logarithmic equations use the esxponents rules to isolate logarithmic expressions with the same base. Set the arguments equal to each other, solve the equation and check your answer.
  • What is logarithm equation?
  • A logarithmic equation is an equation that involves the logarithm of an expression containing a varaible.
  • What are the 3 types of logarithms?
  • The three types of logarithms are common logarithms (base 10), natural logarithms (base e), and logarithms with an arbitrary base.
  • Is log10 and log the same?
  • When there's no base on the log it means the common logarithm which is log base 10.
  • What is the inverse of log in math?
  • The inverse of a log function is an exponantial.

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  • High School Math Solutions – Logarithmic Equation Calculator Logarithmic equations are equations involving logarithms. In this segment we will cover equations with logarithms... Read More

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Logarithmic Word Problems

Log Probs Expo Growth Expo Decay

What are logarithm word problems?

Logarithmic word problems, in my experience, generally involve either evaluating a given logarithmic equation at a given point, or else solving an equation for a given variable; they're pretty straightforward.

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What real-world problems use logarithms?

The classic real-world contexts for logarithm word problems are the measurement of acidity or alkalinity (that is, the measurement of pH), the measurement of sound (in decibels, or dB), and the measurement of earthquake intensity (on the Richter scale), among other uses ( link ).

Note: While log-based word problems are, in my experience, pretty straightforward, their statements tend to be fairly lengthy. Expect to have to plow through an unusual amount of text before they get to the point.

  • Chemists define the acidity or alkalinity of a substance according to the formula pH =  −log[H + ] where [H + ] is the hydrogen ion concentration, measured in moles per liter. Solutions with a pH value of less than 7 are acidic; solutions with a pH value of greater than 7 are basic; solutions with a pH of 7 (such as pure water) are neutral.

a) Suppose that you test apple juice and find that the hydrogen ion concentration is [H + ] = 0.0003 . Find the pH value and determine whether the juice is basic or acidic.

b) You test some ammonia and determine the hydrogen ion concentration to be [H + ] = 1.3 × 10 −9 . Find the pH value and determine whether the ammonia is basic or acidic.

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In each case, I need to evaluate the pH function at the given value of [H + ] . In other words, this exercise, despite all the verbiage, is just plug-n-chug.

Since no base is specified, I will assume that the base for this logarithm is 10 , so that this is the so-called "common" log. (I happen to know that 10 is indeed the correct base, but they should have specified.)

a) In the case of the apple juice, the hydrogen ion concentration is [H + ] = 0.0003 , so:

pH = −log[H + ]

= −log[0.0003]

= 3.52287874528...

This value is less than 7 , so the apple juice is acidic.

b) In the case of the ammonia, the hydrogen ion concentration is [H + ] = 1.3 × 10 −9 , so:

= −log[1.3 × 10 −9 ] = 8.88605664769...

This value is more than 7 , so the ammonia is basic.

(a) The juice is acidic with a pH of about 3.5 , and (b) the ammonia is basic with a pH of about 8.9 .

When a logarithm is given without a base being specified, different people in different contexts will assume different bases; either 10 , 2 , or e . Ask now whether or not bases will be specified for all exercises, or if you're going to be expected to "just know" the bases for certain formulas, or if you're supposed to "just assume" that all logs without a specified base have a base of... [find out which one].

  • "Loudness" is measured in decibels (abbreviated as dB). The formula for the loudness of a sound is given by dB = 10×log[I ÷ I 0 ] where I 0 is the intensity of "threshold sound", or sound that can barely be perceived. Other sounds are defined in terms of how many times more intense they are than threshold sound. For instance, a cat's purr is about 316 times as intense as threshold sound, for a decibel rating of:

dB = 10×log[I ÷ I 0 ]     = 10×log[ (316 I 0 ) ÷ I 0 ]     = 10×log[ 316 ]     = 24.9968708262...

...about 25 decibels.

Considering that prolonged exposure to sounds above 85 decibels can cause hearing damage or loss, and considering that a gunshot from a .22 rimfire rifle has an intensity of about I = (2.5 × 10 13 )I 0 , should you follow the rules and wear ear protection when practicing at the rifle range?

I need to evaluate the decibel equation at I = (2.5 × 10 13 )I 0 :

dB = 10log [ I ÷ I 0 ]     = 10log[ (2.5 ×10 13 )I 0 ÷ I 0 ]     = 10log[2.5 ×10 13 ]     = 133.979400087...

In other words, the squirrel gun creates a noise level of about 134 decibels. Since this is well above the level at which I can suffer hearing damage,

I should follow the rules and wear ear protection.

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  • Earthquake intensity is measured by the Richter scale. The formula for the Richter rating of a given quake is given by R = log[ I ÷ I 0  ] where I 0 is the "threshold quake", or movement that can barely be detected, and the intensity I is given in terms of multiples of that threshold intensity.

You have a seismograph set up at home, and see that there was an event while you were out that had an intensity of I = 989I 0 . Given that a heavy truck rumbling by can cause a microquake with a Richter rating of 3 or 3.5 , and "moderate" quakes have a Richter rating of 4 or more, what was likely the event that occurred while you were out?

To determine the probable event, I need to convert the intensity of the mystery quake into a Richter rating by evaluating the Richter function at I = 989I 0 :

R = log[ I ÷ I 0 ]     = log[ 989I 0 ÷ I 0 ]     = log[989]     = 2.9951962916...

A Richter rating of about 3 is not high enough to have been a moderate quake.

The event was probably just a big truck.

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Introduction

In the CBSE board, chapters of Logarithm are included in the syllabus of classes 9, 10, and 11. Students of class 9 will be introduced to Logarithm questions and answers for the very first time. Hence, the thorough practice of Logarithm problems and answers is the need of the hour. 

However, before proceeding with the chapter on Logarithm, students should be absolutely clear on the basic concepts. It is only then that solving difficult Logarithm questions would become considerably easier. 

Questions Based on Logarithm

Here are some of the Logarithm questions that would impart some ideas to students.  

Question 1: Find Out the Incorrect Statement from Below -

(a) log (1 + 2 + 3) = log 1 + log 2 + log 3

(b) log (2 + 3) = log (2 x 3)

(c) log 10 10 = 1

(d) log 10 1 = 0

Solution: The answer is option (b) log (2 + 3) = log (2 x 3). 

Question 2: What is the Value of log 5 512, when log 2 = 0.3010 and log 3 = 0.4771?

Solution: The answer is option (b) 3.876.

Question 3: Find the Value of log 9, When log 27 Amounts to 1.431.

Solution: An answer is an option (a) 0.954.

Question 4: What is the Value of log 2 10, When log 10 2 = 0.3010?

(a) 1000/301

(b) 699/301

Solution: An answer is an option (a) 1000/301.

Question 5: What is the Value of log 10 80, When log 10 2 = 0.3010?

(d) None of the above option 

Solution: An answer is an option (b) 1.9030. 

Question 6: How Many Digits are there in 2 64 When log 2 = 0.30103?

Solution: The answer is option (b) 20.

Question 7: Which of the Following is True, if a x = b y ?

(a) log a/log b = x/y

(b) log a/b = x/y

(c) log a/log b = y/x

Solution: The answer is option (c) log a/log b = y/x. 

Question 8: What is the Value of log 2 16?

Solution: The answer is (b) 4.

Question 9: Find the Value of y, if log x y = 100 and log 2 x = 10 .

(d) 2 10000

Solution: The answer is option (a) 2 1000 .

Question 10: Find the Value of log 10 (0.0001).

Solution: The answer is option (d) – 4.

Question 11: What is the Value of x When log 2 [log 3 (log 2 x)] = 1?

Solution: The answer is an option (a) 512.

Students’ queries simply and straightforward on Logarithm questions can be clarified in Vedantu’s online classes. You also have the option of downloading PDF materials from the official website. Download the app today!

Logarithm problems and answers are offered for students to solve to fully grasp the subject. These questions are based on the Logarithm chapter in the syllabus for Classes 9, 10, and 11. Practicing these problems will not only help students perform well in academic examinations, but will also allow them to compete in state or national level competitions such as Maths Olympiad.

The Logarithmic function is the exponential function's inverse. It is defined as follows:

If and only if x=ay, then y=logax; for x>0, a>0, and a1.

Logarithms are another way of expressing exponents in Mathematics. A number's Logarithm with a base equals another number. Exponentiation is the inverse function of Logarithm. For instance, if 102 = 100, log10 100 = 2.

As a result, we may deduce that

bn = x or log b x = n

Where b is the Logarithmic function's base.

This may be translated as "the Logarithm of x to the base b equals n."

In this post, we will cover the concept of Logarithms, the two types of Logarithms (common Logarithm and Natural Logarithm), and several Logarithmic characteristics with numerous solved cases.

In the 17th century, John Napier invented the notion of Logarithms. Later, it was utilized by many scientists, navigators, engineers, and others to accomplish numerous computations, which simplified it. Logarithms, in a nutshell, are the inverse process of exponentiation.

What Exactly are Logarithms?

A Logarithm is defined as the number of powers to which a number must be increased to obtain some other numbers. It is the simplest way to express enormous numbers. A Logarithm has many key features that demonstrate that Logarithm multiplication and division may also be stated in the form of Logarithm addition and subtraction.

"The exponent by which b must be raised to give an is the Logarithm of a positive real number a concerning a  base b, a positive real number not equal to 1[nb 1]."

in other words, by=a ⇔logba=y

The letters "a" and "b" represent two positive real numbers.

y is a valid number.

"a" is the argument, which is located within the log, and "b" is the base, which is located at the bottom of the log.

In other terms, the Logarithm answers the question, "How many times is a number multiplied to get the other number?"

For example, how many threes must be multiplied to reach the answer 27?

When we multiply 3 by 3, we get the answer 27.

As a result, the Logarithm is 3.

The Logarithmic form is as follows:

Log3 (27) = 3..... (1)

As a result, the base 3 Logarithm of 27 is three.

The Logarithm form shown above can alternatively be expressed as:

33 = 27..... (2)

As a result, equations (1) and (2) have the same meaning.

Types of Logarithms

In most circumstances, we are dealing with two forms of Logarithms, notably

Common Logarithms

Natural Logarithms

The base 10 Logarithms are another name for the common Logarithm. It is denoted by log10 or just log. The common Logarithm of 1000, for example, is expressed as a log (1000). The common Logarithm specifies how many times the number 10 must be multiplied to obtain the appropriate result.

For instance, log (100) = 2

When we multiply 10 by itself twice, we obtain the number 100.

Natural Logarithm

The base e Logarithm is another name for the natural Logarithm. The natural Logarithm is denoted by the symbols ln or loge. In this case, "e" stands for Euler's constant, which is approximately equal to 2.71828. The natural Logarithm of 78, for example, is represented as ln 78. The natural Logarithm specifies how many times "e" must be multiplied to obtain the appropriate result.

For instance, ln (78) = 4.357.

As a result, the base e Logarithm of 78 is 4.357.

Logarithm Properties and Rules

Logarithmic operations can be conducted according to specific rules. These rules are known as:

Product rule

Division rule

Power rule/Exponential Rule

Change of base rule

Base switch rule

Derivative of log

Integral of log

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FAQs on Logarithm Questions

1. How to Solve Logarithm Basic Questions?

While the solution to be used will vary among logarithmic functions questions, the basic steps involve the following – 

Determining the number of problems present in the logarithm 

Apply relevant properties for simplification of the problem

the problem has to be rewritten sans logarithms

simplify problem further

find the solution of x, and

check the final solution. It must be noted for logs questions that logarithm of a negative number cannot be taken. 

2. What are the Different Properties to Keep in Mind for Solving Log Maths Questions?

There are four properties to be followed for solving logarithm questions. The properties are – (1) product property, (2) quotient property, (3) power property, and (4) change of base property. 

Product rule indicates that multiplying two or more logarithms with common bases becomes equal to the value arising out of separate logarithms. Quotient property lays down that two logarithms having same bases amounts to be equal to result generated from the difference in logarithms. Moreover, in case of change of base property, a given logarithm can be written with a new base. 

3. What are Logarithmic Functions?

The inverse of exponential functions is termed as logarithmic functions. The logarithmic function, y = log a x corresponds to the exponential equation, x = ay. y = log a x. However, this relation holds only under a specific condition. Only if – (1) x = a y , (2) a > 0, and (3) a≠1, will the relation be applicable. 

Having a clear idea about logarithmic functions is essential for solving even the basic log questions.

4. Where can I get good study materials for Logarithms?

Everything you need may be found on the Vedantu app or website. These materials are prepared by experts in the subject, and the information is accurate and dependable. Students will be able to obtain revision notes, important questions, question papers, and much more! There are no fees or costs associated with these study resources. All students need to do is sign in, and then they can download everything they want in pdf format. You may benefit from these free tools, which will undoubtedly help you ace your tests.

5. Is it required to complete all of the exercises provided?

It is always preferable to complete all of the chapter activities because they provide practice. Maths is not a topic that can be learned simply by reading or memorizing. It takes dedication and practice. After each part of the NCERT textbook, there are examples and activities. After you've finished each lesson, check through the solved examples and make sure you understand them properly. Then go to the exercises and attempt to solve them. There is a potential that identical problems will appear in the test, therefore students should practice the sums in the exercises thoroughly.

6. How can I perform well in Math with the help of Vedantu?

The best way to do well in class 9 Maths is to practice. Complete all of the chapter exercises. This will enhance your problem-solving abilities as well as your speed and efficiency. Important formulae, definitions, and equations can be written down in a notebook and reviewed regularly. Solve previous year's sample papers and question papers within a time restriction, all of these resources are available for free on Vedantu. This will familiarise you with the paper design and question style, as well as help you improve your time management abilities. Revise frequently to ensure that you retain all you've learned for a longer amount of time .

7. What exactly are Logarithms and their two types?

Exponents can also be expressed using Logarithms. A Logarithm is defined as the number of powers to which a number must be increased to obtain some other numbers. In other terms, it answers the question, "How many times is one number multiplied to produce the other number?" A number's Logarithm is written as logb x = y. The following are the two most prevalent types of Logarithms:

Base 10 Logarithm (or Common Logarithm)

Base e Logarithm (or Natural Logarithm)

Elon Musk's SpaceX was the satellite villain of the skies. It's taking a 'money hit' to try and fix the problem.

  • SpaceX became astronomers' greatest enemy when it started launching Starlink internet satellites .
  • But SpaceX worked with scientists to try reducing the satellites' brightness in telescope images.
  • Some of SpaceX's solutions sort of work. Other mega-satellite ventures, like Amazon's, are noticing.

Insider Today

NEW ORLEANS — When SpaceX launched its first Starlink satellites , astronomers all over the world freaked out and the company quickly became a villain of the skies.

"I felt as if life as an astronomer and a lover of the night sky would never be the same," astronomer James Lowenthal told the New York Times in 2019.

As the bright trail of satellites climbed through space, ascending to their target orbit in May 2019, people just standing outside could see them clear as day , zipping overhead as they circled Earth.

To some, it seemed to herald the end of astronomy . "If there are lots and lots of bright moving objects in the sky, it tremendously complicates our job," Lowenthal told the Times.

But for rural populations, including in developing countries, that lack reliable internet, Starlink could make a huge difference since SpaceX aims to blanket the Earth in high-speed broadband internet , courtesy of more than 10,000 satellites .

Nonetheless, Starlink satellites — now more than 5,000 strong — are streaking across astronomers' views of the cosmos, ruining their data. Even some telescopes in space aren't safe. Just last year a study found that about one-third of the Hubble Space Telescope's images could be ruined by satellites by 2030.

SpaceX leads the way for change

SpaceX isn't alone in this endeavor; it was just the first company to get huge batches of bright spacecraft off the ground. At least a dozen companies, as well as the Chinese government, are planning to launch their own mega-fleets of satellites .

Many astronomers view the budding business of internet-satellite constellations as an existential threat. But of all the companies racing to claim this new frontier, SpaceX has calmed some of its critics by listening to them, working with them, and trying to darken its satellites.

"Now we're making progress," Jonathan McDowell, an astronomer who has been one of the most outspoken of Starlink's critics , told Business Insider at a conference of the American Astronomical Society in New Orleans.

McDowell and other astronomers were meeting with representatives from the satellite industry to discuss efforts to keep the skies dark and radio-quiet.

No current SpaceX representative participated in the session. Still, Starlink's experimental solutions dominated the conversation — maybe because it's the only company that's been trying the fixes astronomers have suggested.

"For me the focus is not on the call to alarm only, it's on the path to coexistence," Patricia Cooper, a satellite-industry consultant who was formerly VP of SpaceX's government affairs for satellites, told the assembled astronomers. "Not surprisingly, we didn't solve the problem in four and a half years."

SpaceX has tried black paint, sun visors, and now 'mirror film'

SpaceX has thrown a handful of spaghetti at the wall to dim its satellites' shine, and a few things have stuck.

In 2020, about six months after that first bright Starlink trail glided through the skies, SpaceX lobbed its first noodle at the problem by essentially painting a bunch of the satellites black .

That sort of helped. The satellites were less bright, but they were still too bright.

Later that year, SpaceX tried using sun visors to block sunlight from hitting the bottom side of the satellites, where it can reflect back to Earth and make them appear bright.

That worked. The visored satellites were about one-third as bright as those that launched without visors. But they were still bright enough to mess with astronomers' data. The visors were a regular feature for many Starlink satellites until SpaceX added laser communications. The visors blocked the lasers, so they had to go.

Now SpaceX is looking into a "mirror film" that could further reduce brightness on its next Starlink generation. However, those satellites are much bigger than the old ones, "so it kind of cancels out," McDowell said.

"I don't think there are all villain or all hero things here," he added.

SpaceX developed its solutions through meetings with astronomers, including the world's first conference on satellite brightness. In 2022, the International Astronomical Union formalized this ongoing collaboration as the Centre for the Protection of the Dark and Quiet Sky from Satellite Constellation Interference — CPS, for short.

SpaceX has even adopted an operational tweak that astronomers suggested — pointing the satellites' solar panels away from the sun as they pass over the line between night and day. That's when they appear on the horizon, and are most detrimental for telescopes on Earth. Giving the solar panels less sun at this time helps astronomers, but means less energy for the satellite.

"That's real, substantial mitigation that they've done," McDowell said. "They're really taking a money hit by doing that. So we appreciate that."

"If the other companies will do it remains to be seen," he added.

Amazon and other companies may follow SpaceX's lead

Amazon and a small Earth-imaging company called Planet Labs are both following SpaceX's lead.

Chris Hofer, international team lead for Amazon's Project Kuiper internet satellites, told the astronomers in New Orleans that SpaceX's Starlink tinkering has been helpful.

Since joining CPS, Hofer said Amazon has begun improving its solar panels and looking into sun shades.

Both Hofer and Kristina Barkume, of the Earth-imaging satellite company Planet, said they would be following SpaceX's new mirror-film tests with interest.

"Those innovations help us," Barkume told Business Insider.

Though a few companies seem to be paying lip service to, or even throwing money at, the bright-satellite problem, it probably won't go away.

The coming years could see tens of thousands of satellites crowding Earth's orbit. No matter how bright they are or aren't, they will almost certainly interfere with astronomy. Scientists will have to find ways to peep through the gaps between the satellites, however small or fleeting those windows to the cosmos become.

logarithm problems and solutions

Watch: What Elon Musk's 42,000 Starlink satellites could do for — and to — planet Earth

logarithm problems and solutions

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  1. Interesting Logarithmic Equation(2)

  2. Logarithmic equation

  3. INTERESTING LOGARITHMIC EQUATION

  4. #logarithm #propertiesoflogarithms #maths

  5. Logarithmic Equation

  6. PROBLEMS ON LOGARITHM FUNCTIONS# LOGARITHM EQNS & INEQUALITY

COMMENTS

  1. Algebra

    For problems 1 - 3 write the expression in logarithmic form. 75 =16807 7 5 = 16807 Solution 163 4 = 8 16 3 4 = 8 Solution (1 3)−2 = 9 ( 1 3) − 2 = 9 Solution For problems 4 - 6 write the expression in exponential form. log232 = 5 log 2 32 = 5 Solution log1 5 1 625 = 4 log 1 5 1 625 = 4 Solution log9 1 81 = −2 log 9 1 81 = − 2 Solution

  2. Solving Logarithmic Equations

    Example 1: Solve the logarithmic equation. Since we want to transform the left side into a single logarithmic equation, we should use the Product Rule in reverse to condense it. Here is the rule, just in case you forgot. Given Apply Product Rule from Log Rules. Distribute: [latex]\left ( {x + 2} \right)\left ( 3 \right) = 3x + 6 [/latex]

  3. Logarithm Questions

    The questions of logarithm could be solved based on the properties, given below: Product rule: log b MN = log b M + log b N Quotient rule: log b M/N = log b M - log b N Power rule: log b M p = P log b M Zero Exponent Rule: log a 1 = 0 Change of Base Rule: log b (x) = ln x / ln b or log b (x) = log 10 x / log 10 b Also, read: Logarithms

  4. Logarithms

    900 possible mastery points Mastered Proficient Familiar Attempted Not started Quiz Unit test About this unit Logarithms are the inverses of exponents. They allow us to solve challenging exponential equations, and they are a good excuse to dive deeper into the relationship between a function and its inverse. Introduction to logarithms Learn

  5. 4.5e: Exercises

    Answers to odd exercises: B: Basic simplification of logs Exercise 4.5e. B ★ For the following exercises, use properties of logarithms to evaluate without using a calculator. ★ Find a: 39. lna = 1 40. loga = − 1 41. log9a = − 1 42. log12a = 1 43. log2a = 5 44. loga = 13 45. 2a = 7 46. ea = 23

  6. Logarithmic Equations: Problems with Solutions

    Problem 1 Solve the equation \displaystyle \log_2 (x+2)=3 log2(x+2) = 3 Problem 2 Solve the equation \displaystyle \log_9 (3^x)=15 log9(3x) = 15 Problem 3 Solve the logarithmic equation: \displaystyle log_5x=3 log5x = 3 Problem 4 Solve the equation \displaystyle log_x36=2 logx36 = 2 Problem 5

  7. Evaluate logarithms (practice)

    Do 4 problems. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

  8. Logarithmic equations: variable in the argument

    Problems such as these have infinitely many solutions if you allow nonreal complex values of x. Your class most likely won't cover the nonreal complex solutions as that is rather advanced math. So, assuming only real values of x, here is how you would solve a problem like this: 5^(x+3) = 7^(2x+3) ln [ 5^(x+3)] = ln [7^(2x+3) ]

  9. PDF Worksheet: Logarithmic Function

    Vanier College Sec V Mathematics Department of Mathematics 201-015-50 Worksheet: Logarithmic Function 1. Find the value of y. (1) log 5 25 = y (2) log 3 1 = y (3) log 16 4 = y (4) log 2 1 8 = y (5) log

  10. Logarithm Problems with Solutions

    LOGARITHM PROBLEMS WITH SOLUTIONS Problem 1 : Find the logarithm of 64 to the base 2√2. Solution : Write 64 as in terms of 2√2. 64 = 26 = 24+2 = 24 ⋅ 22 = 24 ⋅ [ (√2)2]2 = 24 ⋅ (√2)4 = (2√2)4 log2√264 = log2√2(2√2)4 = 4log2√2(2√2) = 4 (1) = 4 Problem 2 : If logabc = x, logbca = y and logcab = z, then find the value of Solution :

  11. Log Rules Practice Problems With Answers

    Answer Problem 10: Use the rules of logarithms to expand the expression below. Answer You may also be interested in these related math lessons or tutorials: Logarithm Rules Expanding Logarithms Condensing Logarithms Proofs of Logarithm Properties Solving Logarithmic Equations

  12. 4.6e: Exercises

    Exercise 4.6e. 5. ★ For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation. 121. log( 1 100) = − 2. 122. log324(18) = 1 2. ★ For the following exercises, use the definition of a logarithm to solve the equation. 123. 5log7n = 10.

  13. PDF logarithms practice

    Simplify each of the following logarithmic expressions, giving the final answer as a number not involving a logarithm. a) log 24 log 32 2− b) log 96 3log 2 log 43 3 3− − c) 5 5 5 1 2 log 500 log log 10 5 + − d) 2log 54 log 0.25 4log 23 3 3− − e) 8log 2 log 4 3log 96 6 6− −( ) 3 , 1 , 3 , 6 , 6

  14. Logarithms

    This algebra video tutorial provides plenty of practice problems on logarithms including multiple choice problems as well as free response problems.Algebra F...

  15. Logarithmic Equation Calculator

    Set the arguments equal to each other, solve the equation and check your answer. What is logarithm equation? A logarithmic equation is an equation that involves the logarithm of an expression containing a varaible. What are the 3 types of logarithms?

  16. Logarithmic Equations: Very Difficult Problems with Solutions

    Logarithmic Equations: Very Difficult Problems with Solutions. Problem 1. Find the root of the equation \displaystyle 2+lg\sqrt {1+x}+3lg\sqrt {1-x}=lg\sqrt {1-x^2} 2+lg 1 +x +3lg 1−x = lg 1−x2. \displaystyle \frac {9} {100} 1009. \displaystyle \frac {99} {100} 10099. \displaystyle \frac {9} {10} 109. \displaystyle \frac {1} {9} 91.

  17. Algebra

    Here is a set of practice problems to accompany the Solving Logarithm Equations section of the Exponential and Logarithm Functions chapter of the notes for Paul Dawkins Algebra course at Lamar University.

  18. Log problems: pH, decibels, and the Richter Scale

    The classic real-world contexts for logarithm word problems are the measurement of acidity or alkalinity (that is, the measurement of pH), the measurement of sound (in decibels, ... Solutions with a pH value of less than 7 are acidic; solutions with a pH value of greater than 7 are basic; solutions with a pH of 7 (such as pure water) are ...

  19. Logarithm Questions

    Determining the number of problems present in the logarithm Apply relevant properties for simplification of the problem. the problem has to be rewritten sans logarithms. simplify problem further. find the solution of x, and. check the final solution. It must be noted for logs questions that logarithm of a negative number cannot be taken.

  20. Evaluating logarithms (advanced) (video)

    To understand the reason why log(1023) equals approximately 3.0099 we have to look at how logarithms work. Saying log(1023) = 3.009 means 10 to the power of 3.009 equals 1023. The ten is known as the base of the logarithm, and when there is no base, the default is 10. 10^3 equals 1000, so it makes sense that to get 1023 you have to put 10 to ...

  21. What the border bill would and wouldn't do

    At a time when problems at the border often grab headlines, the compromise proposed in the Senate offers a rare list of suggested solutions with a bipartisan bent.

  22. Environmental ethics, sustainability and decisions: literature problems

    Environmental ethics, sustainability and decisions: literature problems and suggested solutions by Fabio Zagonari, Springer, 2022, xxxv + 253 pp., €114 (hardcover book), €93 (ebook), ISBN 978-3-031-21181-2

  23. Calculus I

    Section 1.9 : Exponential And Logarithm Equations. For problems 1 - 12 find all the solutions to the given equation. If there is no solution to the equation clearly explain why. 12−4e7+3x = 7 12 − 4 e 7 + 3 x = 7 Solution. 1 = 10−3ez2−2z 1 = 10 − 3 e z 2 − 2 z Solution. 2t−te6t−1 = 0 2 t − t e 6 t − 1 = 0 Solution.

  24. SpaceX leads the way for change

    SpaceX's Starlink internet satellites brighten the night sky enough to interfere with scientific research. It's testing solutions to appease critics.