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 maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

If $\displaystyle 5x^{log _23} + 3^{log _2x} = 162$ then logarithm of $x$ to the base 4 has the value equal to :

  1. $2$
  2. $1$
  3. $-1$
  4. $3/2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given,

$5x^{\log _2\left(3\right)}+3^{\log _2\left(x\right)}=162$

$3^{\log _2\left(x\right)}=162-5x^{\log _2\left(3\right)}$

$\Rightarrow \log _2\left(3^{\log _2\left(x\right)}\right)=\log _2\left(162-5x^{\log _2\left(3\right)}\right)$

$\Rightarrow \log _2\left(x\right)\log _2\left(3\right)=\log _2\left(162-5x^{\log _2\left(3\right)}\right)$

$\Rightarrow \log _2\left(x^{\log _2\left(3\right)}\right)=\log _2\left(162-5x^{\log _2\left(3\right)}\right)$

$x^{\log _2\left(3\right)}=162-5x^{\log _2\left(3\right)}$

$x=27^{\frac{665}{1054}}\approx 8$

Now,

$\log _4x$

$=\log _4 8$

$=\dfrac{3}{2}$
Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

The value of $ a^{\frac{\log _b (\log _b N)}{\log _b a}}$ is

  1. $\log _b (N-b)$
  2. $\log _b (N+b)$
  3. $\log _b\dfrac Nb$
  4. $\log _b N$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Using $\log _p q =\dfrac{\log q}{\log p}$
$\displaystyle= a^{\cfrac{\log _b (\log _b N)}{\log _b a}}$
$\displaystyle = a^{\cfrac{\log(\log _b N)/\log b}{\log a/\log b}}$
$=\displaystyle a^{\displaystyle \log _a (\log _b N)}$
$=\log _b N[\because a^{\log _ab=b}]$


Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

If ${ log } _{ 4 }5=a\quad and\quad { log } _{ 5 }6=b,\quad then\quad { log } _{ 3 }2$ is equal to

  1. $\dfrac { 1 }{ 2a+1 } $
  2. $\dfrac { 1 }{ 2b+1 } $
  3. $2ab+1$
  4. $\dfrac { 1 }{ 2ab-1 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given,

$\log _{4}5=a$

$\log _{5}6=b$

now,

$ab=\log _{4}5 \times \log _{5}6$

$=\dfrac{\log 5}{\log 4} \times \dfrac{\log 6}{\log 5}$

$=\dfrac{\log 6}{\log 4}$

$ab=\log _4{6}$

$=\dfrac{1}{2}\log _2 6$

$=\dfrac{1}{2}\log _2 (2\times 3)$

$=\dfrac{1}{2}(\log _2 2+\log _2 3)$

$=\dfrac{1}{2}(1+\log _2 3)$

$2ab=1+\log _2 3$

$\log _2 3=2ab-1$

$\therefore \log _3 2=\dfrac{1}{2ab-1}$

Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

If $4^{\log _{2}\log x}=\log x-\left ( \log x \right )^{2}+1$ (base is e), then find the value of $x$

  1. $x=e$
  2. $x=2e$
  3. $x=3e$

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

Given equation is 
$4^{\log _{2}\log x}=\log x-\left ( \log x \right )^{2}+1$

Now, $\log _{2}\log x$ is meaningful if $\log x > 0$.
Since $4^{\log _{2}\log x}=2^{2\log _{2}\log x}=\left ( 2^{\log _{2}\log x} \right )^{2}=\left ( (\log x)^{\log _{2}2} \right )^{2}[\because a^{\log _bc}=c^{\log _ba}]$
             $=\left ( \log x \right )^{2}$    $ \left ( \because a^{\log _{a}x}=x, a> 0, a\neq 1 \right )$

So the given equation reduces to
$2\left ( \log x \right )^{2}-\log x-1=0$.
$\displaystyle \Rightarrow \log x=1, \log x=-\frac{1}{2}$.
But $\log x> 0$
Hence,    $\log x=1$, i.e.,$x=e$

Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

The value of $\left( \log _{ b }{ a }  \right) \left( \log _{ c }{ b }  \right) \left( \log _{ a }{ c }  \right) $ is equal to

  1. $0$
  2. $\log { abc } $
  3. $1$
  4. $10$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $x=\left( \log _{ b }{ a }  \right) \left( \log _{ c }{ b }  \right) \left( \log _{ a }{ c }  \right) \ \Rightarrow x=\left( \dfrac { \log { a }  }{ \log { b }  }  \right) \left( \dfrac { \log { b }  }{ \log { c }  }  \right) \left( \dfrac { \log { c }  }{ \log { a }  }  \right),\left[\because \log _yx=\cfrac{\log x}{\log y}\right] \ \Rightarrow x=1$

Ans: C

Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

Sometimes to solve an equation, we may use the identity ${ a }^{ \log _{ a }{ b }  }=b,b>0,a>0,a\neq 1$
Then solution set of $3{ x }^{ \log _{ 5 }{ 2 }  }+{ 2 }^{ \log _{ 5 }{ x }  }=64$ is,

  1. $5$
  2. $25$
  3. $125$
  4. $625$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$3{ x }^{ \log _{ 5 }{ 2 }  }+{ 2 }^{ \log _{ 5 }{ x }  }=64$


Since, ${ a }^{ \log _{ b }{ c }  }={ c }^{ \log _{ b }{ a }  }$

Therefore, $3\left( { 2 }^{ \log _{ 5 }{ x }  } \right) +{ 2 }^{ \log _{ 5 }{ x }  }=64$

$\Rightarrow 4 \cdot 2^{\log _5x}=64$

$\Rightarrow { 2 }^{ \log _{ 5 }{ x }  }=16={ 2 }^{ 4 }$

Equating power of 2

$\Rightarrow \log _{ 5 }{ x } =4$

$\Rightarrow x={ 5 }^{ 4 }=625$

Ans: D

Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

Sometimes to solve an equation, we may use the identity ${ a }^{ \log _{ a }{ b }  }=b,b>0,a>0,a\neq 1$
Then the number of solution(s) of ${ x }^{ \log _{ x }{ { \left( x+3 \right)  }^{ 2 } }  }=16$ is/are,

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

  4. infinite

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

${ x }^{ \log _{ x }{ { \left( x+3 \right)  }^{ 2 } }  }=16$


Above equation is valid when $x>0,x\neq 1$

${ x }^{ \log _{ x }{ { \left( x+3 \right)  }^{ 2 } }  }=16$
     
$\Rightarrow (x+3)^2=16=4^2, [\because { a }^{ \log _{ a }{ b }  }=b]$

$\Rightarrow x+3=\pm 4$

$\Rightarrow x=1,-7$

Since, $x>0,x\neq 1$

Therefore, $x=\left{ \emptyset  \right} $

Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

Using the identity $\displaystyle a^{\log _{a}{n}}= n,$ find:
$\displaystyle 3^{-\tfrac{1}{2}\log _{3}9}$, 

  1. $0.33$
  2. $-0.33$
  3. $0.66$
  4. $-0.66$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given $a^{log _a n}=n$    ....(1)

Consider, $\displaystyle 3^{-\frac{1}{2}\log _{3}9}$
$= 3^{\log _{3}9^{-1/2}}$
$= 9^{-1/2}$        (by (1))
$= \displaystyle \frac{1}{3}$
$=0.33$

Ans: 0.33
Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

Using the identity $\displaystyle a^{\log _{a}{n}}= n,$ find:

$\displaystyle 2^{2-\log _{2}5}$

  1. $0.6$
  2. $0.8$
  3. $0.2$
  4. $0.1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given $\displaystyle a^{\log _{a}{n}}= n$      .....(1)

Consider, 

$\displaystyle 2^{2-\log _2 5}=$$\displaystyle 2^{2}.2^{-\log _{2}5}$

                    $= 4.2^{\log _{2}5^{-1}}$

                    $= 4\,.5^{-1}$       $(by (1))$

                    $= \cfrac{4}{5}$

$\displaystyle 2^{2-\log _2 5}=0.8$

Ans: 0.8
Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

If $\displaystyle a^{\log _{a}10}= 10$, then the set of value(s) of $a$ is/are

  1. $\displaystyle a \in \left ( 0,1 \right )\cup \left ( 1,\infty \right )$
  2. $\displaystyle a \in \left [ 0,1 \right )\cup \left (1,\infty \right).$
  3. $\displaystyle a \in \left ( -1,0 \right )\cup \left ( 1,\infty \right )$
  4. $\displaystyle a \in \left (-1,0\right ]\cup \left ( 1,\infty \right ).$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle a^{\log _{a}10}= 10$
$\displaystyle \log _a10.\log _{10}a = 1\Rightarrow \cfrac{\log _{10}a}{\log _{10}a}=1$
Given expression is correct for all $'a'$ provided $'a'$ belongs to the domain of the given expression.
$\displaystyle \log _{a}10$ is defined for a+ive, and $\displaystyle a\neq 1.$
$\displaystyle \therefore a \in \left ( 0,1 \right )\cup \left ( 1,\infty  \right)$

Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

If $\displaystyle \log _{p}q+\log _{q}r+\log _{r}p$ vanishes, where $p,q$ and $r$ are positive reals different than unity, then the value of $\displaystyle \left ( \log _{p}q \right )^{3}+\left ( \log _{q}r \right )^{3}+\left ( \log _{r}p \right )^{3} $ is

  1. an odd prime.

  2. an even number.

  3. an odd composite.

  4. an irrational number.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given 
$\displaystyle \log _{p}q+\log _{q}r+\log _{r}p=0$
$\Rightarrow \log _{p}q+\log _{q}r=-\log _{r}p$    ....(1)

Consider,$\left( \log _{ p } q \right) ^{ 3 }+\left( \log _{ q } r \right) ^{ 3 }+\left( \log _{ r } p \right) ^{ 3 }$

$=[(\log _{ p } q+\log _{ q } r)^{ 3 }-3\log _{ p } q\log _{ q } r(\log _{ p } q+\log _{ q } r)]+\left( \log _{ r } p \right) ^{ 3 }$

$=\left( -\log _{ r } p \right) ^{ 3 }-3\log _{ p } q\log _{ q } r\left( -\log _{ r } p \right) +\left( \log _{ r } p \right) ^{ 3 }$       (by (1))

$=3\log _{ p } q\log _{ q } r\log _{ r } p = 3 \quad \quad [\because \log _ab=\dfrac{\log b}{\log a}]$

Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

The value of ${\left(\displaystyle\frac{1}{2}\right)}^{\log _{2}5}$ is equal to

  1. $ \displaystyle\frac{1}{5}$
  2. $ \displaystyle\frac{-1}{5}$
  3. $ \displaystyle\frac{-1}{25}$
  4. $ \displaystyle\frac{1}{25}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle { \frac { 1 }{ 2 }  }^{ \log _{ 2 }{ 5 }  }={ 2 }^{ -\log _{ 2 }{ 5 }  }={ 2 }^{ \log _{ 2 }{ { 5 }^{ -1 } }  }={ 5 }^{ -1 }[\because a^{\log _ab}=b]=\frac { 1 }{ 5 } $

Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

The value of the expression
$\displaystyle\frac{1}{1+\log _b\,a+\log _b\,c}+\displaystyle\frac{1}{1+\log _c\,a+\log _c\,b}+\displaystyle\frac{1}{1+\log _a\,b+\log _a\,c}$ is equal to

  1. $\,\,abc$
  2. $\,\,\displaystyle\frac{1}{abc}$
  3. $\,\,0$
  4. $\,\,1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle\frac{1}{1+\log _b\,a+\log _b\,c}+\displaystyle\frac{1}{1+\log _c\,a+\log _c\,b}+\displaystyle\frac{1}{1+\log _a\,b+\log _a\,c}$
$=\displaystyle \dfrac{1}{\displaystyle 1+\frac{\log a}{\log b}+\frac{\log c}{\log b}}+\dfrac{1}{\displaystyle 1+\frac{\log a}{\log c}+\frac{\log b}{\log c}}+\dfrac{1}{\displaystyle 1+\frac{\log b}{\log a}+\frac{\log c}{\log a}} [\because \log _yx=\dfrac{\log x}{\log y}]$
$=\displaystyle\frac{\log\,b}{\log\,b+\log\,a+\log\,c}+\displaystyle\frac{\log\,c}{\log\,c+\log\,a+\log\,b}+\displaystyle\frac{\log\,a}{\log\,c+\log\,a+\log\,b}\,=\,1$

Ans: D

Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

The value of $\,3^{\textstyle \log _4\,5}\,+\,4^{\textstyle \log _5\,3}\,-5^{\textstyle \log _4\,3}\,-3^{\textstyle \log _5\,4}$ is equal to

  1. $\,\,0$
  2. $\,\,1$
  3. $\,\,2$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Use $\,a^{\textstyle \log _b\,c}\,=\,c^{\textstyle \log _b\,a}$
So, $[5^{\textstyle \log _4\,3}=3^{\textstyle \log _4\,5}]$
and $[4^{\textstyle \log _5\,3}=3^{\textstyle \log _5\,4}]$
$\Rightarrow\,\,3^{\textstyle \log _4\,5}\,+\,4^{\textstyle \log _5\,3}\,-\,3^{\textstyle \log _4\,5}\,-\,4^{\textstyle \log _5\,3}\,=\,0$

Ans: A