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 parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

If $y = \log \left( \frac { 1 + x } { 1 - x } \right) ^ { 1 / 4 } - \frac { 1 } { 2 } \tan ^ { - 1 } x ,$ then $\frac { d y } { d x } =$

  1. $\frac { x ^ { 2 } } { 1 - x ^ { 4 } }$
  2. $\frac {2 x ^ { 2 } } { 1 - x ^ { 4 } }$
  3. $\frac { x ^ { 2 } } { 2 \left( 1 - x ^ { 4 } \right) }$

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

$\dfrac{dy}{dx}=\dfrac{1}{(\dfrac{1+x}{1-x})^{\frac{1}{4}}}\times$ $\dfrac{1}{4\times{(\dfrac{1+x}{1-x})^{\frac{3}{4}}}}\times $ $\dfrac{(1-x)(1)-(1+x)(-1)}{(1-x)^{2}}-$ $\dfrac{1}{2}\dfrac{1}{x^2+1}$ 


$\dfrac{1-x}{4(1+x)}\times\dfrac{2}{(1-x)^2}$ $-\dfrac{1}{2(x^2+1)}$  $=\dfrac{1}{2(1-x^2)}-$ $\dfrac{1}{2(x^2+1)}$ =$\dfrac{x^2}{1-x^4}$

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

$\displaystyle \log _{10}x + \log _{10}y \geq 2$, then the smallest possible value of $\displaystyle x + y$ is

  1. $\displaystyle 10$
  2. $\displaystyle 30$
  3. $\displaystyle 20$

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

${log} _{10}x+{log} _{10} y=$ $log _{10}(xy)\geq2$, 
Thus, $xy\geq100$
 Given the product of two numbers ,addition of two number is smallest when they are equal.
$ x^2\geq100$
Therefore, smallest value of $x+y =20$
Hence, option 'C' is correct.

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

The solution set of the system of equations $\log _{ 3 }{ x } +\log _{ 3 }{ y } =2+\log _{ 3 }{ 2 } \quad and\quad \log _{ 27 }{ (x+y) } =\dfrac { 2 }{ 3 } $ is :

  1. {6,3}

  2. {3,6}

  3. {6,12}

  4. {12,6}
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,

$\log _3\left(x\right)+\log _3\left(y\right)=2+\log _3\left(2\right)$

$\log _3\left(x\right)=2+\log _3\left(2\right)-\log _3\left(y\right)$

$x=3^{2+\log _3\left(2\right)-\log _3\left(y\right)}$

$=3^{\log _3\left(2\right)}\cdot \:3^{-\log _3\left(y\right)}\cdot \:3^2$

$=2\cdot \:3^{-\log _3\left(y\right)}\cdot \:3^2$

$=2y^{-1}\cdot \:3^2$

$\Rightarrow x=\dfrac{18}{y}$.......(1)

Now,

$\log _{27}\left(x+y\right)=\dfrac{2}{3}$

from (1)

$\log _{27}\left(\dfrac{18}{y}+y\right)=\dfrac{2}{3}$

$\dfrac{18}{y}+y=27^{\frac{2}{3}}$

$\Rightarrow 18+y^2=9y$

$y^2-9y+18=0$

$(y-3)(y-6)=0$

$\therefore y=3,6$

$x=\dfrac{18}{y}=\dfrac{18}{3}=6$

$x=\dfrac{18}{y}=\dfrac{18}{6}=3$

$(x,y)=\left \{ 6,3 \right \}or\left \{ 3,6 \right \}$
Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

Number of real solutions of the equation $\sqrt { \log _{ 10 }{ (-x) }  } =\log _{ 10 }{ \sqrt { { x }^{ 2 } }  } $ is :

  1. zero

  2. exactly 1

  3. exactly 2

  4. 5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,

$\sqrt{\log _{10}(-x)}=\log _{10}(\sqrt{x^2})$

$\sqrt{\log _{10}(-x)}=\log _{10}x$

$\log _{10}\left(-x\right)=\left(\log _{10}\left(x\right)\right)^2$

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

let $\log _{10}\left(x\right)=u$

$\log _{10}\left(-1\right)+u=\left(u\right)^2........\log _{10}(-1) \ is \ not \ defined$

$u=\mathrm{Undefined}$

$\:\log _{10}\left(x\right)=\mathrm{Undefined}:\quad x=10^{\mathrm{Undefined}}$

$x=10^{\mathrm{Undefined}}\space\mathrm{False}$

No solution for $\:x\in \mathbb{R}$
Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

Number of ordered pair(s) of (x,y) satisfying the system of equations, $\log _2 xy = 5$ and $\log _{\frac{1}{2}} \frac{x}{y} = 1$ is:

  1. one

  2. two

  3. three

  4. four
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,

$\log _2\left(xy\right)=5,\:\log _{\frac{1}{2}}\left(\frac{x}{y}\right)=1$

$\log _2\left(xy\right)=5$

$xy=2^5=32$

$x=\dfrac{32}{y}$

$\log _{\frac{1}{2}}\left(\dfrac{x}{y}\right)=1$

$\log _{\frac{1}{2}}\left(\dfrac{\frac{32}{y}}{y}\right)=1$

$\dfrac{\frac{32}{y}}{y}=\left(\dfrac{1}{2}\right)^1$

$\dfrac{32}{y}=\dfrac{1}{2}y$

$y^2=64$

$\Rightarrow y=\pm 8$

$x=\dfrac{32}{\pm 8}=\pm 4$

$(4,8),(-4,-8)$ Therefore $2$ ordered pairs
Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

If $\log _3{(\log _3{a})}+\log _{\cfrac{1}{3}}{\left(\log _{\cfrac{1}{3}}{b}\right)}=1$, then the value of $ab^3$ is 

  1. $9$
  2. $3$
  3. $1$
  4. $\cfrac{1}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let x = log3(a) and y = log(1/3)(b). The equation is log3(x) + log(1/3)(y) = 1. Since log(1/3)(y) = -log3(y), we have log3(x) - log3(y) = 1, so log3(x/y) = 1, meaning x/y = 3, or x = 3y. Substituting back: log3(a) = 3 * log(1/3)(b) = 3 * (-log3(b)) = -3 * log3(b) = log3(b^-3). Thus, a = b^-3, which means a * b^3 = 1.

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

The value of $\log _a n\times\log _n m $  is equal to

  1. $\log _a m$
  2. $\log _m a$
  3. $\dfrac{\ln m}{\ln a}$
  4. $\dfrac{\ln a}{\ln m}$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$\because \displaystyle \log _{ n }{ m } =\frac { \log{ m }  }{ \log { n }  } $
$\therefore \log _an\times \log _nm=\dfrac{\log n}{\log a}\times \dfrac{\log m}{\log n}=\dfrac{\log m}{\log a}=\log _am$

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

If $x = \displaystyle \frac{y}{(1 + x)^p}$, then $p$ is equal to

  1. $\displaystyle \frac{\displaystyle \log _e \left ( \frac{y}{x} \right )}{\log _e (1 + a)}$
  2. $\log \displaystyle \left \{ \frac{y}{x(1+ a)} \right \}$
  3. $\log \displaystyle \left \{ \frac{y - x}{1+ a} \right \}$
  4. $\displaystyle \frac{\log y}{\log \{ x(1 + a) \}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since $x = \displaystyle \frac{y}{(1 + a)^p}$


$\therefore   (1 + a)^p = \displaystyle \frac{y}{x}$
or $p   \log _e (1 + a) = \log _e\dfrac{y}{x}$
or $\displaystyle p = \dfrac{\displaystyle log _e\left ( \frac{y}{x} \right )}{log _e (1 + a)}$

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

The value of $\log _{ 2 }{ 7 } $ is:

  1. an integer

  2. a prime number

  3. a rational number

  4. an irrational number
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Suppose $\log _{2}{7}$ is rational.

$\Rightarrow \: \log _{2}{7}=\dfrac{a}{b}\:\Rightarrow \: 7=2^{a/b}$
$\Rightarrow \: 7^{b}=2^{a}$

But $2^{a}$ is even and $7^{b}$ is odd.
Hence, our assumption is wrong.

$\Rightarrow \: \log _{2}{7}$ is irrational.
Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

if $y=\left( \log _{ 2 }{ 3 }  \right) \left( \log _{ 3 }{ 4 }  \right) ....\left( \log _{ 31 }{ 32 }  \right) $, then

  1. $4< y\le 5$
  2. $y=5$
  3. $4< y< 6$
  4. $y=6$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

Let $y = (\log _2 3 ) (\log _3 4 ) ( \log _4 5 ) … (\log _{31} 32 )$

Then, $2^y =2^{((\log _2 3 ) (\log _3 4 ) ( \log _4 5 ) … (\log _{31} 32 ))} $

By laws of exponents and the definition of a logarithm,

$2^{((\log _2 3 ) (\log _3 4 ) ( \log _4 5 ) … (log _{31} 32 ))} $

$=(2^{(\log _23)})^{((\log _3 4 ) ( \log _4 5 ) … (\log _{31} 32 ))} $

$= 3^{((\log _3 4 ) ( \log _4 5 ) … (\log _{31} 32 ))} $

$=(3^{(\log _34)})^{(( \log _4 5 ) … (\log _{31} 32 ))} $

$=4^{(( \log _4 5 ) … (\log _{31} 32 ))} $ .......

$=31^{(\log _{31}32)} =32$

$\therefore 2^y =32$

$y=5$