Mathematics · Quantitative Aptitude

Logarithms

246 Questions

Logarithms are mathematical operations that determine the exponent required for a base to reach a specific number. This topic tests the application of logarithmic properties, changing bases, and solving complex equations. It is a high-yield topic for quantitative aptitude in competitive exams.

Logarithmic expressionsBase change propertiesSolving log equationsInfinite series logsCharacteristic values

Logarithms Questions

Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

The value of x, which satisfies the equation $2 \log _ { 2 } \left( \log _ { 2 } x \right) + \log _ { 12 } \left( \log _ { 2 } ( 2 \sqrt { 2 } x ) \right) = 1$ is greater

  1. 10

  2. 11

  3. 7

  4. 9
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving the logarithmic equation requires using properties of logs and substitution. The resulting value of x satisfies the inequality x > 10.

Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

The logarithm form of $\displaystyle 5^3 = 125$ is equal to

  1. $\displaystyle \log _5 125 = 3$
  2. $\displaystyle \log _5 125 = 5$
  3. $\displaystyle \log _3 125 = 5$
  4. $\displaystyle \log _5 3 = 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$5^3=125$
Taking log on both sides, we get
$3\log 5 = \log 125$
$\log _5 125 = 3$         ...(since $\dfrac{\log a}{\log b} = \log _ba$)

Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

The logarithmic form of $\displaystyle (81)^{\frac {3}{4}} = 27$ is

  1. $\displaystyle \log _{66} 36 = \frac {2}{9}$
  2. $\displaystyle \log _{81} 27 = \frac {3}{4}$
  3. $\displaystyle \log _{16} 33 = \frac {7}{2}$
  4. $\displaystyle \log _{78} 12 = \frac {1}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$81^{\frac{3}{4}}=27$
Taking log on both sides
$\dfrac{3}{4}log81=log27$
$log _{81}27= \dfrac{3}{4}$.....(since $\dfrac{loga}{logb}=log _ba$)

Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

$\log V = 2 \log 2 - \log 3 + \log \pi + 3 \log r$ can be expressed as

  1. $V = \dfrac{4}{3} \pi r^{3}$
  2. $ V = \dfrac{2}{3} \pi r^{3}$
  3. $ V = \dfrac{4}{3} \pi r$
  4. $ V = \dfrac{2}{3} \pi r$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$ \log { V }  = 2\log { 2 } -\log { 3 } +\log { \pi  }  + 3\log { r }$
$ \log V = \log ({ 2 }^{ 2 }\times \pi \times { r }^{ 3 }) - \log { 3 }$
$ \log V = \log \dfrac { 4\pi { r }^{ 3 } }{ 3 } $
Removing log
$V = \dfrac {4}{3} \pi r^3$
Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

The value of $7 log _a \displaystyle \frac{16}{15} + 5 log _a \frac{25}{24} + 3 log _a \frac{81}{80}$ is

  1. $log _{a3}$
  2. $log _{a1}$
  3. $log _{a2}$
  4. $log _{a5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have,

$7log _a\dfrac{16}{15}+5log _a\dfrac{25}{24}+3log _a\dfrac{81}{80}$
$\Rightarrow log _a(\dfrac{16}{15})^7+log _a(\dfrac{25}{24})^5+log _a(\dfrac{81}{80})^3$
$\Rightarrow log _a(\dfrac{16}{15})^7\times (\dfrac{25}{24})^5\times (\dfrac{81}{80})^3$
$\Rightarrow log _a\dfrac{16^3\times 16^4}{5^7\times 3^7}\times \dfrac{5^5\times 5^5}{8^5\times 3^5}\times \dfrac{3^6\times 3^6}{16^3\times 5^3}$
$\Rightarrow log _a\dfrac{16^4}{1\times 1}\times \dfrac{1}{8^5\times 1}\times \dfrac{1}{1}$
$\Rightarrow log _a\dfrac{2^4\times 8^4}{8^5}$
$\Rightarrow log _a\dfrac{16}{8}$
$\Rightarrow log _a2$

Hence, this is the answer.

Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

If $log _{10} x - log _{10} \sqrt x = \displaystyle \frac{2}{log _{10} x}$, then value of x is

  1. $\displaystyle \frac{1}{100}$ or $100$
  2. $\pm$ 2
  3. 10 or $\displaystyle \frac{1}{10}$

  4. 100
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider the given equation.

$log _{10}x-log _{10}\sqrt x=\dfrac{2}{log _{10}x}$
$log _{10}\dfrac{x}{\sqrt x}=\dfrac{2}{log _{10}x}$
$log _{10}\sqrt x=\dfrac{2}{log _{10}x}$
$\dfrac{1}{2}log _{10}\ x=\dfrac{2}{log _{10}x}$
$\dfrac{1}{2}(log _{10}\ x)^2=2$
$(log _{10}\ x)^2=4$
$log _{10}\ x=\pm 2$
$x=10^{\pm2}$
$x=100\ or\ \dfrac{1}{100}$

Hence, this is the answer.

Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

If $\displaystyle \frac{log _2 (9 - 2^x)}{3 - x} = 1$, then value of x is

  1. x = 4

  2. x = + 1 or -1
  3. x = $\pm$ 2

  4. x = 0
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have,

$\dfrac{log _2(9-2^x)}{3-x}=1$
$log _2(9-2^x)=3-x$
$9-2^x=2^{3-x}$                $ .......... (1)$

From option $(D)$
$9-2^0=2^{3-0}$
$9-1=2^3$
$8=8$

Hence, $x=0$ is the root of this equation.

Hence, only option $D$ is correct.

Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

The equation  ${ \left( \log _{ 10 }{ x+2 }  \right)  }^{ 3 }+{ \left( \log _{ 10 }{ x-1 }  \right)  }^{ 3 }={ \left( 2\log _{ 10 }{ x+1 }  \right)  }^{ 3 }$ has

  1. no natural solution

  2. two rational solutions

  3. no prime solution

  4. one irrational solution
Reveal answer Fill a bubble to check yourself
B,C,D Correct answer
Explanation

Let $ \log _{ 10 }{ x+2 } =a$ and $ \log _{ 10 }{ x-1 } =b$
$\therefore a+b=2\log _{ 10 }{ x+1 } $ (from the question)
Thus, the given equation(in the question) reduces to ${a}^{3}+{b}^{3}={(a+b)}^{3}$
$\Rightarrow 3ab(a+b)=0$
$\Rightarrow a=0$ or $b=0$ or $a+b=0$
$\Rightarrow \log _{ 10 }{ x+2 }=0$  or $\log _{ 10 }{ x-1 }=0$ or $2\log _{ 10 }{ x } +1=0$
$\Rightarrow x={10}^{-2}$  or  $x=10$ or  $x={ 10 }^{ -\frac { 1 }{ 2 }  }$
Hence  $x=\left{ \dfrac { 1 }{ 100 },10 ,\dfrac { 1 }{ \sqrt { 10 }  }  \right} $