Mathematics · Quantitative Aptitude

Logarithms

231 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 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

$x$ and $y$ are real numbers such that ${7^x} - 16y = 0\;{\text{and}}\;{4^x} - 49y = 0,$ then the value of $\left( {y - x} \right)$ is

  1. $\dfrac{5}{2}$
  2. $\dfrac{{19}}{5}$
  3. $\dfrac{{4115}}{{2013}}$
  4. $\dfrac{{1569}}{{784}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$7^{x}=16y$

$4^{x}=49 y$

$\Rightarrow \dfrac{7^{x}}{4^{x}} = \dfrac{16}{49}$

$\Rightarrow \left( \dfrac{7}{4} \right)^{x} = \left( \dfrac{4}{7} \right)^{2} = \left( \dfrac{7}{4} \right)^{-2}$

$\Rightarrow x=-2$

$y= \dfrac{7^{x}}{16}$

$\Rightarrow y= \dfrac{1}{49 \times 16}$

So, $y-x = \dfrac{1}{49 \times 16}+2$

$=\dfrac{1}{784}+2$

$=\dfrac{1569}{784}$
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.