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 logarithm and its uses basic mathematical concepts physics

The value of $x$ which satisfy $log(x+1) = 2logx$ is 

  1. $1$
  2. $\dfrac{\sqrt{5}-1}{2}$
  3. $\dfrac{\sqrt{5}+1}{2}$
  4. $2$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

$\log(x+1)=2\log x$


$\Rightarrow \log (x+1)=\log x^2$

$\Rightarrow x^2-x-1=0$

$\Rightarrow x=\cfrac{+1\pm \sqrt{1+4}}{2}=\cfrac{1\pm \sqrt{5}}{2}$

$\Rightarrow x= \left(\cfrac{1+\sqrt{5}}{2}\right),\left(\cfrac{1-\sqrt{5}}{2}\right)$

Multiple choice logarithm and its uses basic mathematical concepts physics

The value of $x$ satisfying the equation $g^{log _3 (log _2 x)} = log _2 x - (log _2 x)^2 + 1$ is 

  1. $0$
  2. $1$
  3. $2$

  4. None
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, $g^{\log _{3}(\log _{2}x)}$ is an exponential function and $\log _{2}{x}-\left(\log _{2}{x}\right)^{2}+1$ is a quadratic with imaginary roots.

The two can be equal when both side become $0,1$. Since, right hand side can become zero at imaginary point. We, only consider, then the two side become $1$.
$\log _{2}{x}-\left(\log _{2}{x}\right)^{2}+1=1$
$\Rightarrow \left(\log _{2}{x}\right)^{2}-\left(\log _{2}{x}\right)=0$
$\Rightarrow \left(\log _{2}{x}\right)\left(\log _{2}{x}-1\right)=0$
$\Rightarrow \log _{2}{x}=0$ and $\log _{2}{x}=1$
$\Rightarrow x=1,2$
But $x\neq 1$ as in $g^{\log _{3}(\log _{2}x)}$ it become invalid hence, $x=2$ satisfy the relation.

Multiple choice logarithm and its uses basic mathematical concepts physics

If $2y = log(12-5x-3x^2)$ takes all real values then $x$ belongs to 

  1. $(-3, 5/3)$
  2. $(-3, 3)$
  3. $(-3, 4/3)$

  4. None
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$2y=\log\left(12-5x-3x^2\right)$

$(12-5x-3x^2)>0$
$3x^2+5x-12<0$
$3x^2+9x-4x-12<0$
$3x\left(x+3\right)-4\left(x+3\right)<0$
$\left(x+3\right)\left(3x-4\right)<0$
$x\epsilon \left(-3,{4/3}\right)$

Multiple choice logarithm and its uses basic mathematical concepts physics

Evaluate the expression by using logarithm tables: $ \dfrac{(17.42)^{2/{3}}\times 18.42}{\sqrt{126.37}}$

  1. $11.01$
  2. $12.01$
  3. $13.01$
  4. $14.01$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $x= \dfrac{(17.42)^{2/{3}}\times 18.42}{\sqrt{126.37}}$
Taking logarithm on both sides,
$ \log { x } =\log { (17.42)^{ 2/{ 3 } } } +\log { 18.42 } -\log { \sqrt { 126.37 }  } $

$\log { x } =\dfrac { 2 }{ 3 } \log { 17.42 } +\log { 18.42 } -\dfrac { 1 }{ 2 } \log { (126.37) } $

$ \log { x } =\dfrac { 2 }{ 3 } \log { (1.742\times 10) } +\log { (1.842\times 10) } -\dfrac { 1 }{ 2 } \log { (1.264\times { 10 }^{ 2 }) } $

$ \log { x } =\dfrac { 2 }{ 3 } { (1.2410) } + { 1.2653 } -\dfrac { 1 }{ 2 } { (2.1018) } $

$\log { x } =0.8273+1.2653-1.0509$

$\log { x } =1.0417$
$\Rightarrow x= \text{antilog }(1.0417)$
$\Rightarrow x = 11.01$

Multiple choice logarithm and its uses basic mathematical concepts physics

Let $a = \log 3\log _32$. An integer k satisfying  $1< 2^{(-k+3^{-a})} < 2,$  must be less than ____.

  1. $1.25766$
  2. $2.256$
  3. $3$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given a = log3(log3(2)). The inequality 1 < 2^(-k + 3^-a) < 2 simplifies to 0 < -k + 3^-a < 1. Since 3^-a = 3^-log3(log3(2)) = 1/log3(2) = log2(3) approx 1.585. So 0 < -k + 1.585 < 1, which means 0.585 < k < 1.585. Thus k must be less than 1.585.

Multiple choice logarithm and its uses basic mathematical concepts physics

If $a=\log _35 $ and $b= \log _725$ then correct option is:

  1. $a < b$
  2. $ a > b$
  3. $a= b$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$a=\log _{ 3 }{ 5 } =\cfrac { \log { 5 }  }{ \log { 3 }  } ,b=\cfrac { \log { 25 }  }{ \log { 7 }  } =\cfrac { \log { 5^2 }  }{ \log { 7 }  }=\cfrac { 2\log { 5 }  }{ \log { 7 }  } $


$ \cfrac { a }{ b } =\cfrac { \log { 5 }  }{ \log { 3 }  } \times \cfrac { \log { 7 }  }{ 2\log { 5 }  } =\cfrac { 1 }{ 2 } \log _{ 3 }{ 7 } =\log _{ 3 }{ (\sqrt { 7 } ) } $


$ Now,\sqrt { 7 } <3,so\quad \cfrac { a }{ b } <1$

$ \cfrac { a }{ b } <1\  =>a<b$

Multiple choice logarithm and its uses basic mathematical concepts physics

The value of ${ \left( 0.05 \right)  }^{ \log _{ \sqrt { 20 }  }{ \left( 0.1+0.01+0.001+.... \right)  }  }$ is 

  1. $81$
  2. $\cfrac{1}{81}$
  3. $20$
  4. $\cfrac{1}{20}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$0.1+0.01+0.0001+..... \Rightarrow G.P$
$=(0.1)(1-0.1)\quad S _{\infty}=-\dfrac {a}{1-r}$
$=\dfrac {0.1}{0.9}=\dfrac {1}{9}$
$\therefore \ (0.05)\log _\sqrt {20} (0.1+0.01+....)$
$=\left (\dfrac {1}{20}\right)\log \sqrt {20}^{1/9}$
$=\left (\dfrac {1}{9}\right) \log \sqrt {20}^{1/20}$
$=\left (\dfrac {1}{20}\right) \log \sqrt {20}^{(\sqrt {20})^{-2}}=\left (\dfrac {1}{20}\right)^{-2}$
$=81$
Multiple choice logarithm and its uses basic mathematical concepts physics

If $\log _{10}e=0.4343$, then $\log _{10}1016$ is

  1. $2.99$
  2. $3$
  3. $3.006949$
  4. $3.02$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\log _{10}1016\Rightarrow \dfrac{\log 1016}{\log 10}$

$\Rightarrow \dfrac{3.006949}{1}$
$\rightarrow$ Option $C$ is correct

Multiple choice logarithm and its uses basic mathematical concepts physics

Multiple Correct:

Which of the following statements are true

  1. $\log _{ 2 }{ 3 } <\log _{ 12 }{ 10 } $
  2. $\log _{ 6 }{ 5 } <\log _{ 7 }{ 8 } $
  3. $\log _{ 3 }{ 26 } <\log _{ 2 }{ 9 } $
  4. $\log _{ 16 }{ 15 } >\log _{ 10 }{ 11 } >\log _{ 7 }{ 6 }$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Multiple choice logarithm and its uses basic mathematical concepts physics

The solution of the equation $\log _{7}\log _{5}(\sqrt {x^{2}}+5+x)=0$

  1. $x=2$
  2. $x=3$
  3. $x=0$
  4. $x=-2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\log _{7}\log _{5}(\sqrt{x^{2}}+5+x)=0$

$\log _{5}(\sqrt{x^{2}}+5+x)=1$
$\sqrt{x^{2}+5+x=5}$
$\sqrt{x^{2}+x=0}$
$x^{2}+x^{2}+2\sqrt[x]{x^{2}}=0$
$x=0$


Multiple choice logarithm and its uses basic mathematical concepts physics

Find the value of $\log _{10}{\left(0.\bar{9}\right)}$

  1. $0$
  2. $1$
  3. $-1$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
To find value of $\log _{10}{0.\bar9}$

Let $x=0.\bar{9}=0.999999...$

$\Rightarrow 10x=9.99999....$

$\Rightarrow 10x-x=9$

$\Rightarrow 9x=9$

$\Rightarrow x=\dfrac{9}{9}=1$

$\therefore x=1$

Let $y=\log _{10}{1}$

$\Rightarrow 1={10}^{y}$

$\Rightarrow {10}^{y}={10}^{0}$                 (since ${10}^{0}=1$)

Since bases are same we can equate the powers

$\therefore y=0$

Hence, $\log _{10}{\left(0.\bar{9}\right)}=0$