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

If $log _x \left( \dfrac{9}{16} \right) = - \dfrac{1}{2}$, then x is equal to

  1. $- \dfrac{3}{4}$
  2. $\dfrac{3}{4}$
  3. $\dfrac{81}{256}$
  4. $\dfrac{256}{81}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$log _x \left( \dfrac{9}{16} \right ) = - \dfrac{1}{2}$
$\Rightarrow x^{-1/2} = \dfrac{9}{16}$
$\Rightarrow \dfrac{1}{\sqrt x} = \dfrac{9}{16}$
$\Rightarrow \sqrt x = \dfrac{16}{9}$
$\Rightarrow x = \left( \dfrac{16}{9} \right)^2$
$\Rightarrow x = \dfrac{256}{81}$

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

What is the value of $\dfrac {1}{2}\log _{10} 25 - 2 \log _{10} 3 +\log _{10} 18$?

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

The value of $\dfrac {1}{2}\log _{10} 25 - 2 \log _{10} 3 +\log _{10} 18$ is
$= \log _{10}(25)^{1/2} - \log _{10} (3)^{2} + \log _{10}18$
$= \log _{10}5 - \log _{10}9 + \log _{10}18$
$= \log _{10} \left (\dfrac {5}{9}\times 18\right ) $

$= \log _{10} 10 $    ....Using the identity $\log _aa=1$
$= 1$

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

The logarithmic form of ${5}^{2}=25$ is

  1. $\log _{ 5 }{ 2 } =25$
  2. $\log _{ 2 }{ 5 } =25$
  3. $\log _{ 5 }{ 25 } =2$
  4. $\log _{ 25 }{ 5 } =2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$5^2=25$

Taking log with base $5$ both sides, we get
$\log _55^2=\log _525$
$\Rightarrow \log _525=2\log _55$
$\Rightarrow \log _525=2$     $(\log _aa=1)$
Hence, C is the correct option.

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

If $2\log y -\log x -3=0$, express $x$ in terms of $y.$

  1. $x=\dfrac{y^2}{e^3}$
  2. $x=\dfrac{y^2}{e^2}$
  3. $x^2=\dfrac{y^2}{e^3}$
  4. $x=\dfrac{y^3}{e^3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\Rightarrow$$\log { { y }^{ 2 } } -\log { x } -\log { { e }^{ 3 } } =0$.......$\log e=1$

$\Rightarrow$$ \log { x } =\log { \left (\cfrac { { y }^{ 2 } }{ { e }^{ 3 } } \right ) } $

$\Rightarrow$$ x=\cfrac { { y }^{ 2 } }{ { e }^{ 3 } } $

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

If $2\log y -\log x-3=0$ express $x$ in terms of $y.$

  1. $x^2=1000y$
  2. $x^2= \dfrac{y^2}{e^3}$
  3. $y^2= \dfrac{x}{1000}$
  4. $y^2= 1000x$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given: $2\log y -\log x-3=0$

$\log { { y }^{ 2 } } -\log { x } -3\log { { e }=0 } $.......$(\log e=1)$

$\log { { y }^{ 2 } } -\log { x } -\log { { e }^{ 3 }=0 } $

$ \log { x } =\log { { y }^{ 2 } } -\log { { e }^{ 3 } } =\log { \left (\cfrac { { y }^{ 2 } }{ { e }^{ 3 } } \right ) } $

$ x=\cfrac { { y }^{ 2 } }{ { e }^{ 3 } } $
Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

If $2x^{{log _4}^3}+3^{\log _4x}=27$, then x is equal to?

  1. $2$
  2. $4$
  3. $8$
  4. $16$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$2\times { x }^{ { log } _{ 4 }3 }+{ 3 }^{ { log } _{ 4 }x }=27$

${ 2\times3 }^{ { log } _{ 4 }x }+{ 3 }^{ { log } _{ 4 }x }=27$

Let ${ 3 }^{ { log } _{ 4 }x }=t$
$2t+t=27$
$3t=27$
$t=9$

${ 3 }^{ { log } _{ 4 }^{ x } }={ 3 }^{ 2 }=9$

So,  ${ log } _{ 4 }x=2$

So,  $x={ 2 }^{ 4 }=16$
Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

The value of x, for which the 6th term in the expansion of $\left{ { 2 }^{ { log } _{ 2 }\sqrt { \left( { 9 }^{ x-1 }+7 \right)  }  }+\dfrac { 1 }{ { 2 }^{ { \left( 1/5 \right) log } _{ 2 }\left( { 3 }^{ x-1 }+1 \right)  } }  \right} ^{ 7 }$ is 84, is equal to 

  1. 4

  2. 3

  3. 2

  4. 1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The 6th term of the binomial expansion (a+b)^7 is 7C5 * a^2 * b^5. Solving the equation for x leads to x=2.

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

If x = ${ log } _{ 3 }243,y={ log } _{ 2 }64,$, Then $\sqrt { x-2\sqrt { y }  } $ is 

  1. $\sqrt { 5-2\sqrt6 }$
  2. $2-\sqrt { 3 } $
  3. $\sqrt { 3 } -\sqrt { 2 } $
  4. $\sqrt { 3 } -4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have,

$ x={{\log } _{3}}243\,\,......\,\,\left( 1 \right) $

$ y={{\log } _{2}}64\,\,......\,\,\left( 2 \right) $

Applying rule,

${{\log } _{b}}x=y\,\Rightarrow {{b}^{y}}=x$

So,

From equation (1) and (2) to,

$ x={{\log } _{3}}243 $

$ \Rightarrow {{3}^{x}}=243 $

$ \Rightarrow {{3}^{x}}={{3}^{5}} $

$ \Rightarrow x=5 $

Now,

$ {{\log } _{2}}64=y $

$ \Rightarrow {{2}^{y}}=64 $

$ \Rightarrow {{2}^{y}}={{2}^{6}} $

$ \Rightarrow y=6 $

Now,

$ \sqrt{x-2\sqrt{y}} $

$ =\sqrt{5-2\sqrt{6}} $

Hence, this is the answer.

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

Logarithmic form of  $3 \sqrt { 8 } = 2$  is

  1. $\log _ { 8 } 2 = \dfrac { 1 } { 3 }$
  2. $\log _ { 2 } 8 = \dfrac { 1 } { 3 }$
  3. $\log _ { \frac { 1 } { 3 } } 8 = 2$
  4. $\log _ { \frac { 1 } { 3 } } 2 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The exponential form 3*sqrt(8) = 2 is not standard. However, if the expression is 8^(1/3) = 2, then the logarithmic form is log_8(2) = 1/3. The provided option A matches this logic.

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

Number of solutions of $\log _{4}{\left(x-1\right)}=\log _{2}{\left(x-3\right)}$

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\log _{4}{\left(x-1\right)}=\log _{2}{\left(x-3\right)}$
$\Rightarrow\,\log _{{2}^{2}}{\left(x-1\right)}=\log _{2}{\left(x-3\right)}$
$\Rightarrow\,\dfrac{1}{2}\log _{2}{\left(x-1\right)}=\log _{2}{\left(x-3\right)}$
$\Rightarrow\,\log _{2}{\left(x-1\right)}=2\log _{2}{\left(x-3\right)}$
$\Rightarrow\,\left(x-1\right)={\left(x-3\right)}^{2}$
$\Rightarrow \,x-1={x}^{2}-6x+9$
$\Rightarrow \,{x}^{2}-6x-x+9+1=0$
$\Rightarrow \,{x}^{2}-7x+10=0$
$\Rightarrow \,{x}^{2}-2x-5x+10=0$
$\Rightarrow \,x\left(x-2\right)-5\left(x-2\right)=0$
$\Rightarrow \,\left(x-2\right)\left(x-5\right)=0$
$\therefore\,x=2,\,5$
Number of solutions$=2$