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 $\dfrac{log _2 24}{log _{96} 2}-\dfrac{log _2192}{log _{12}{2}}$ is

  1. $3$
  2. $0$
  3. $2$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Consider
$\dfrac{log _2 24}{log _{96} 2}-\dfrac{log _2192}{log _{12}{2}}\\$
$=\dfrac{log24.log96-log192log12}{(log2)^2}$
$=\dfrac{log(2^3 \times 3)log(2^5\times 3)-log(2^6\times3)log(2^2\times3)}{(log2)^2}$
$=\dfrac{(3log2+log3)(5log2+log3)-(6log2+log3)(2log2+log3)}{(log2)^2}$
$=\dfrac{15(log2)^2-12(log2)^2}{(log2)^2}$
$=3\dfrac{(log2)}{log2}$
$=3$
Option A is the correct answer.
Multiple choice logarithm and its uses basic mathematical concepts physics

The greatest value of $(4\log _{10}{x}-\log _{2}{(0.0001)})$ for $0 < x < 1$ is

  1. $4$
  2. $-4$
  3. $8$
  4. $-8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expression is 4*log10(x) - log2(10^-4) = 4*log10(x) + 4*log2(10). This does not have a simple maximum for 0 < x < 1 as it approaches -infinity. The question is likely garbled.

Multiple choice logarithm and its uses basic mathematical concepts physics

If $P$ is the number of natural numbers whose logarithm to the base $10$ have the characteristic $p$ and $Q$ is the number of natural numbers logarithm of whose reciprocals to the base $10$ have the characteristic $-q$, then find the value of $\log _{10}P-\log _{10}Q$.

  1. $p-q+1$
  2. $p+q-1$
  3. $p+q$
  4. $p-q$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $x$ and $y$ be the numbers whose logarithm to the base $10$ have the characteristic $p$ and $q$ respectively.
$10^{p}\leq x< 10^{p+1}\Rightarrow P=10^{p+1}-10^{p}\Rightarrow P=9\times 10^{p}$
Similarly, $10^{q-1}< y\leq 10^{q}$
$\Rightarrow $   $Q=10^{q}-10^{q-1}=10^{q-1}\left ( 10-1 \right )=9\times 10^{q-1}$
$\therefore $   $\log _{10}P-\log _{10}Q=\log _{10}\left ( P/Q \right )=\log _{10}10^{p-q+1}=p-q+1$

Multiple choice logarithm and its uses basic mathematical concepts physics

Find the number of positive integers which have the characteristic $3$, when the base of the logarithm is $7$.

  1. $2058$
  2. $1029$
  3. $1030$
  4. $2060$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let there be $N$ integers whose characteristic is 3, when base of log is 7
Then, $\log _{ 7 }{ N } =x$ where $3\le x<4$
As $3\le x<4$
$3\le \log _{ 7 }{ N } <4\ \Rightarrow { 7 }^{ 3 }\le N<{ 7 }^{ 4 }\ \Rightarrow N={ 7 }^{ 4 }-{ 7 }^{ 3 }\ \Rightarrow N=2401-343=2058$ 

Multiple choice logarithm and its uses basic mathematical concepts physics

The value of $\displaystyle anti\log _5\left [\frac {\tan^2\left (\frac {\pi}{5}\right )+\tan^2\left (\frac {2\pi}{5}\right )+20}{\cot^2\left (\frac {\pi}{5}\right )+\cot^2\left (\frac {2\pi}{5}\right )+28}\right ]$ is equal to

  1. odd number

  2. even number

  3. prime number

  4. composite number
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation
$\frac { { tan }^{ 2 }(\frac { \Pi  }{ 5 } )+{ tan }^{ 2 }(\frac { 2\Pi  }{ 5 } )+20 }{ { cot }^{ 2 }(\frac { \Pi  }{ 5 } )+{ cot }^{ 2 }(\frac { 2\Pi  }{ 5 } )+28 } \\ =\frac { { (\sqrt { 5-2\sqrt { 5 }  } ) }^{ 2 }+(\sqrt { \frac { 5 }{ 5-2\sqrt { 5 }  }  } )^{ 2 }+20 }{ { (\frac { 1 }{ \sqrt { 5-2\sqrt { 5 }  }  } ) }^{ 2 }+{ (\sqrt { \frac { 5-2\sqrt { 5 }  }{ 5 } ) }  }^{ 2 }+28 } \\ =\frac { 5-2\sqrt { 5 } +\frac { 5 }{ 5-2\sqrt { 5 }  } +20 }{ \frac { 1 }{ 5-2\sqrt { 5 }  } +\frac { 5-2\sqrt { 5 }  }{ 5 } +28 } \\ =\frac { 5[(5-{ 2\sqrt { 5 } ) }^{ 2 }+5+20(5-2\sqrt { 5 } )] }{ 5+(5-{ 2\sqrt { 5 } ) }^{ 2 }+140(5-2\sqrt { 5 } ) } \\ =\frac { 5(45-20\sqrt { 5 } +5+100-40\sqrt { 5 } ) }{ 5+45-20\sqrt { 5 } +700-200\sqrt { 5 }  } \\ =\frac { 750-300\sqrt { 5 }  }{ 750-300\sqrt { 5 }  } \\ antilog _{ 5 }[\frac { { tan }^{ 2 }(\frac { \Pi  }{ 5 } +{ tan }^{ 2 }(\frac { 2\Pi  }{ 5 } )+20 }{ { cot }^{ 2 }(\frac { \Pi  }{ 5 } )+{ cot }^{ 2 }(\frac { 2\Pi  }{ 5 } )+28 } ]={ 5 }^{ 1 }\quad =5$
Multiple choice logarithm and its uses basic mathematical concepts physics

Evaluate using logarithm table: $\dfrac {28.45 \times \sqrt [3] {0.3254}}{32.43 \times \sqrt [5] {0.3046}}$

  1. $0.7666$
  2. $0.7656$
  3. $0.5686$
  4. $0.2936$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $y=\dfrac { 28.45\times \sqrt [ 3 ]{ .3254 }  }{ 32.43\times \sqrt [ 3 ]{ .3046 }  } $

$ \ln { y } =\ln { 28.45 } +\ln { \sqrt [ 3 ]{ .3254 }  } -(\ln { 32.43 } +\ln { \sqrt [ 5 ]{ .3046 }  } )\ \ln { y } =\ln { 25.45 } +\dfrac { 1 }{ 3 } \ln { .3245- } \ln { 32.43 } -\dfrac { 1 }{ 5 } \ln { .4046 } \ \ln { y } =3.236+(-.375)-3.479-(-.237)\ \ln { y } =-.381$
$ y=$ anti $\ln { (-.381) } $

$ y=.7656$
So, option B is correct.

Multiple choice logarithm and its uses basic mathematical concepts physics

If $\log _{10} 2 = 0.3010$, then the number of digits in $2^{64}$ is

  1. $18$
  2. $24$
  3. $22$
  4. $20$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given $\log _{ 10 }{ 2 } =0.301$

$\log _{ 10 }{ 2^{64} } =64 \times \log _{ 10 }{ 2 } =64 \times 0.3010=19.264$
$\Rightarrow 2^{64}=10^{19.264}$
The number of digits in $10^{19}$ is $20$ , there will be $21$ digits from $10^{21}$
The number $10^{19.264}$ lies between them
Therefore the number of digits in $10^{19.264}$ is $20$
Therefore the correct option is $D$

Multiple choice logarithm and its uses basic mathematical concepts physics

If $\log _{10} 3 = 0.4771$, then the number of zeros after the decimal in $3^{-100}$ is

  1. $47$
  2. $48$
  3. $49$
  4. $50$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The number of zeroes will be
$=|log _{10}(3^{-100})|$
$=|-100(log _{10}(3))|$
$=|-47.71|$
$=47.71$
Taking the integral part (since number of zeroes has to be an integer, there will be $47$ zeros.

Multiple choice logarithm and its uses basic mathematical concepts physics

Approximate of $\log _{11}21$ is

  1. 1.27

  2. 1.21

  3. 1.18

  4. 1.15

  5. 1.02
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Approximate value of $\log _{ 11 }{ 21 } $

$=\log _{ 11 }{ (7\times 3) } $
$=\log _{ 11 }{ 7 } +\log _{ 11 }{ 3 }$
$ =0.8115+0.4581$
$=1.27$

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

Vant Hoff's equation is ___.

  1. ${log\frac{K _2}{K _1}=\frac{-\Delta H^{0}}{2.303R}\left [ \frac{T _2-T _1}{T _2T _1} \right ]}$.
  2. ${log\frac{K _2}{K _1}=\frac{\Delta H^{0}}{2.303R}\left [ \frac{T _2-T _1}{T _2+T _1} \right ]}$.
  3. ${log\frac{K _2}{K _1}=\frac{\Delta H^{0}}{2.303R}\left [ \frac{T _2-T _1}{T _2T _1} \right ]}$.
  4. ${log\frac{K _2}{K _1}=\frac{\Delta H^{0}}{2.303R}\left [ \frac{T _2+T _1}{T _2T _1} \right ]}$.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
The van't Hoff equation provides information about the temperature dependence of the equilibrium constant. The van't Hoff equation may be derived from the Gibbs-Helmholtz equation, which gives the temperature dependence of the Gibbs free energy.

The van't Hoff equation is $K=Ae^{\Delta H/RT}$ or $\displaystyle\frac {d ln K}{\partial T}=\frac {\Delta H}{RT^2}$

By, integrating the above equation, you will get the required relation.

Hence, the given statement is correct
Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

If $\displaystyle \log _{16} 8$ = $\displaystyle \frac {3}{m}$, then value of $m$ is equal to 

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\log _{16}8=\dfrac{3}{m}$

$\therefore \dfrac{\log 8}{\log 16}=\dfrac{3}{m}$      ....($\log _ba=\dfrac{\log a}{\log b}$)

$\therefore \dfrac{\log 2^3}{\log 2^4}=\dfrac{3}{m}$   ...($\log a^b=a\log b$)

$\therefore \dfrac{3}{m}=\dfrac{3}{4}$ $ \Rightarrow m = 4$.
Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If $log2,log({ 2 }^{ x }-1)and\quad log({ 2 }^{ x }+3)$ are in A.P., then x is equal to :

  1. $\dfrac { 5 }{ 2 } $
  2. ${ log } _{ 2 }5$
  3. ${ log } _{ 3 }2$
  4. ${ log } _{ 5 }2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\log { 2 } ,\log { \left( { 2 }^{ x }-1 \right)  } ,\log { \left( { 2 }^{ x }-3 \right)  } $ are in AP

$\log { \left( { 2 }^{ x }-1 \right)  } =\log { 2 } +\log { \left( { 2 }^{ x }-3 \right)  } $
${ \log { \left( { 2 }^{ x }-1 \right)  }  }^{ 2 }=\log { \left[ 2.\left( { 2 }^{ x }+3 \right)  \right]  } $
$\quad { \left( { 2 }^{ x }-1 \right)  }^{ 2 }={ 2 }^{ x+1 }+6$
${ \left( { 2 }^{ x } \right)  }^{ 2 }+1-2.{ 2 }^{ x }=2.{ 2 }^{ x }+6$
${ \left( { 2 }^{ x } \right)  }^{ 2 }-4.{ 2 }^{ x }-5=0\Rightarrow { \left( { 2 }^{ x } \right)  }^{ 2 }-5.{ 2 }^{ x }+{ 2 }^{ x }-5=0\Rightarrow { 2 }^{ x }({ 2 }^{ x }-5)+1({ 2 }^{ x }-5)=0\Rightarrow ({ 2 }^{ x }-5)({ 2 }^{ x }+1)=0$
${ 2 }^{ x }+1\neq 0,{ 2 }^{ x }+5=0\Rightarrow { 2 }^{ x }=5$
taking log of base 2
$x\log _{ 2 }{ 2 } =\log _{ 2 }{ 5 } $
$x=\log _{ 2 }{ 5 } $

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If $1,\,{\log _y}x,\,{\log _z}y,\, - \,15{\log _{x}z}$ are in $A.P.$ , then  

  1. ${z^3} = x$
  2. $x = {y^{ - 1}}$
  3. ${z^{ - 3}} = y$
  4. $x = {y^{ - 1}} = {z^3}$

  5. All the above
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Let $d$ be the common difference of the $A.P.$

Then,
$\log _yx=1+d$
$\Rightarrow$  $x=y^{1+d}$                     ----- ( 1 )

$\log _zy=1+2d$
$\Rightarrow$  $y=z^{1+2d}$                   ------ ( 2 )

$-15\log _xz=1+3d$
$\Rightarrow$  $z=x^{\frac{-(1+3d)}{15}}$             ------ ( 3 )

$x=y^{1+d}=z^{(1+2d)(1+d)}=x^{\tfrac{-(1+d)(1+2d)(1+3d)}{15}}$

$\Rightarrow$  $(1+d)(1+2d)(1+3d)=-15$

$\Rightarrow$  $6d^3+11d^2+6d+16=0$

$\Rightarrow$  $(d+2)(6d^2-d+8)=0$

$\Rightarrow$  $d=-2$

Substituting value of $d$ we get,

$\Rightarrow$  $x=y^{-1}=z^3=x^{\tfrac{1}{3}}$ or

$\Rightarrow$  $x=y^{-1}=z^3,\,y=z^{-3}$