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 sum to infinite terms of a gp sequence, progression and series maths

Value of $y = {\left( {0.64} \right)^{{{\log } _{0.25}}\left( {\cfrac{1}{3} + \cfrac{1}{{{3^2}}} + \cfrac{1}{{{3^3}}}....upto   \infty } \right)}}$ is :

  1. $0.9$
  2. $0.8$
  3. $0.6$
  4. $0.25$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$y= (0.64)^{log _{0.25} \left(\dfrac{1}{3}+ \dfrac{1}{3^{2}}+ \dfrac{1}{3^{3}}+..... \right)}$
$=(0.64)^{\log _{0.25}^{\left( \dfrac{\dfrac{1}{3}}{1-1/3} \right)}}$
$=(0.64)^{\log _{0.25} } \left( \dfrac{1}{2} \right)$
$= (0.64)^{\log 0.5} _{0.25}$
$(0.64)^{0.5}= (0.64)^{1/2}= \sqrt{0.64}= 0.8$
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $x>0$ and $\displaystyle log _{2}x+log _{2}(\sqrt{x})+log _{2} (\sqrt[4]{x})+log _{2}(\sqrt[8]{x})+...\infty =4 ,$then $x=$

  1. 2

  2. 3

  3. 4

  4. 5
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $log _{2}x+log _{2}(\sqrt{x})+log _{2} (\sqrt[4]{x})+log _{2}(\sqrt[8]{x})+...\infty =4 $

$\Rightarrow log _{2}[x.x^{1/2}.x^{1/4}.x^{1/8}...\infty

]=log _{2}[x^{1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...\infty} ]=log _{2}

x^{\dfrac{1}{1-(1/2)}}=log _{2}(x^{2})=4 $

$ \therefore x^{2}=2^{4}=16

 \therefore x=4$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

The value of $a^{\log _{2}}x$, where $a=0.2,b=\sqrt {5},x=\dfrac {1}{4}+\dfrac {1}{8}+\dfrac {1}{16}+.....$ to $\infty $ is

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

$x=\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 8 } +\cfrac { 1 }{ 16 } +....,a=0.2,b\sqrt { 5 } \ x\cfrac { \cfrac { 1 }{ 4 }  }{ 1-\cfrac { 1 }{ 2 }  } \ x=\cfrac { 1 }{ 2 } \ { a }^{ \log _{ b }{ x }  }=(0.2)^{ \log _{ \sqrt { 5 }  }{ \cfrac { 1 }{ 2 }  }  }\ =(\cfrac { 1 }{ 2 } )^{ \log _{ \sqrt { 5 }  }{ 0.2 }  }\ =(\cfrac { 1 }{ 2 } )^{ -2 }\ a^{ \log _{ b }{ x }  }=4$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $A = 1 + {r^a} + {r^{2a}} + {r^{3a}}......\infty $ and $B = 1 + {r^b} + {r^{2b}}......\infty$ then$\dfrac{a}{b} = $

  1. $\dfrac{\log{\left({A-1}\right)}}{\log{\left({B-1}\right)}}$
  2. $\dfrac{\log{\left(\dfrac{A-1}{A}\right)}}{\log{\left(\dfrac{B-1}{B}\right)}}$
  3. $\dfrac{\log{\left({A}\right)}}{\log{\left({B}\right)}}$
  4. $\dfrac{\log{\left({B}\right)}}{\log{\left({A}\right)}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If $\left|Common\, ratio\right|<1$, then sum of infinite terms of G.P. is

$S=\dfrac{First\,  Term}{1\,−\,Common \,Ratio}$

$A=1+{r}^{a}+{r}^{2a}+{r}^{3a}+...$

Common ratio$=\dfrac{{r}^{a}}{1}={r}^{a}$

$A=\dfrac{1}{1−{r}^{a}}$

$\Rightarrow\,1−{r}^{a}=\dfrac{1}{A}$

$\Rightarrow\,{r}^{a}=1−\dfrac{1}{A}$

$\Rightarrow\,{r}^{a}=\dfrac{A−1}{A}$

Take $\log$ on both

$\log{\left({r}^{a}\right)}=\log{\left(\dfrac{A−1}{A}\right)}$

As $\log{\left({m}^{n}\right)}=n\log{\left(m\right)}$

$\Rightarrow\,a\log{\left(r\right)}=\log{\left(\dfrac{A−1}{A}\right)}$  ....$(i)$

$B=1+{r}^{b}+{r}^{2b}+{r}^{3b}+...$

Common ratio$=\dfrac{{r}^{b}}{1}={r}^{b}$

$B=\dfrac{1}{1−{r}^{b}}$

$\Rightarrow\,1−{r}^{b}=\dfrac{1}{B}$

$\Rightarrow\,{r}^{b}=1−\dfrac{1}{B}$

$\Rightarrow\,{r}^{b}=\dfrac{B−1}{B}$

Take $\log$ on both

$\log{\left({r}^{b}\right)}=\log{\left(\dfrac{B−1}{B}\right)}$

As $\log{\left({m}^{n}\right)}=n\log{\left(m\right)}$

$\Rightarrow\,b\log{\left(r\right)}=\log{\left(\dfrac{B−1}{B}\right)}$  ....$(ii)$

$\dfrac{(i)}{(ii)}=\dfrac{a\log{r}}{b\log{r}}=\dfrac{\log{\left(\dfrac{A-1}{A}\right)}}{\log{\left(\dfrac{B-1}{B}\right)}}$

$\therefore\,\dfrac{a}{b}=\dfrac{\log{\left(\dfrac{A-1}{A}\right)}}{\log{\left(\dfrac{B-1}{B}\right)}}$
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

The sum of the series
$\dfrac { 1 } { 1.2 } - \dfrac { 1 } { 2.3 } + \dfrac { 1 } { 3.4 } \ldots \ldots \ldots$  up to  $\infty$  is equal to

  1. $\log _{ { { e } } } \left( \dfrac { 4 }{ { e } } \right) $
  2. $2 \log _ { e } 2$
  3. $\log _ { e } 2 - 1$
  4. $\log _ { e } 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

lf $e^{(\cos^{2}x+\cos^{4}x+\cos^{6}x+\ldots.)\log 3}$ satisfies $y^{ 2 }-10y+9=0$ and $0\le x\le \cfrac { \pi  }{ 2 } $, then $\cot^{2}x=$

  1. $0$
  2. $1$
  3. $\dfrac12$
  4. $9$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle e^{(\cos^{2}x+\cos^{4}x+\cdots )\log{3}}=e^{\left(\dfrac{\cos^{2}x}{1-\cos^{2}x}\right)\log3.}$
$\because (\cos^{2}x+\cos^{4}x+\cdots)$ is forming an infinite G.P. $=e^{\cot^{2}x\log3.}$
$y^{2}-10y+9=0  \Rightarrow   y=9,1$
$e^{\cot^{2}x\log{3}}=9,1$
$3^{\cot^{2}x}=9,1&gt;&gt;[\because e ^{\log x}=x]$
$\Rightarrow \cot^{2}x=2,0$ but as $x\in \left[0,\dfrac{\pi }{2}\right]$
$\Rightarrow \cot^{2}x=0$

Hence, option 'A' is correct.

Multiple choice logarithm and its uses basic mathematical concepts physics

If $\log _3 x = 3\, &amp; \,\log _x y = 4\,$, then find $y$.

  1. ${3^6}$
  2. ${3^9}$
  3. ${3}^{12}$
  4. $none$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have,

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

 

We know that

${{\log } _{a}}x=y\Rightarrow x={{a}^{y}}$

 

Therefore,

$ x={{3}^{3}} $

$ x=27 $

 

Here,

$ {{\log } _{x}}y=12 $

$ y={{x}^{12}} $

 

On putting the value of $x$, we get

$y={{3}^{12}}$

 

Hence, this is the answer.

Multiple choice logarithm and its uses basic mathematical concepts physics

If anti ${ \log } _{ 10 }(0.3678)=2.3324$ then ${ \log } _{ 10 }233.2$ is equal to

  1. 367.8

  2. 36.78

  3. 3.3678

  4. 2.3678
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$antilog _{10}(0.3678)=2.3324$


$\therefore \log(2.3324)=0.3678$

$\log _{10}233.2=\log _{10}(2.332\times 100)$

                     $=\log _{10}(2.332) + \log (10^2) \quad \dots ( \log a+\log b=\log(ab))$

                     $=\log _{10}(2.332) + 2\log (10) \quad \dots ( n \log _ab=\log _ab^n)$

As we know, $log _a a=1$

$\log _{10}233.2=2+0.3678=2.3678$

Multiple choice logarithm and its uses basic mathematical concepts physics

Which of the following real numbers is(are) non-positive?

  1. $log{ } _{ 0.3 }(\dfrac { \sqrt { 5 } +2 }{ \sqrt { 5 } -2 } )$
  2. $log{ } _{ 7 }(\sqrt { 83 } -9\quad )$
  3. $log{ } _{ 7\frac { \pi }{ 12 } }(cot\frac { \pi }{ 8 } \quad )$
  4. ${ log } _{ 2 }\sqrt { 9.\sqrt [ 3 ]{ { 27 }^{ \frac { -5 }{ 3 } }.243{ }^{ \frac { -7 }{ 5 } } } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
From given,

we have,

$\log _{0.3}\left(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\right)$

$=\log _{0.3}\left(2+\sqrt{5}\right)-\log _{0.3}\left(\sqrt{5}-2\right)$

$=-2.39811$
Multiple choice logarithm and its uses basic mathematical concepts physics

If $\log x = -2.0258$, then $x$ is equal to

  1. $0.009223$
  2. $0.009423$
  3. $0.008422$
  4. $0.008223$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\log x=-2.0258$

Characteristics$=-3$
Mantissa$=-2.0258-(-3)=0.9742$
Value of $0.9742$ from antilog table $=9419+4=9423$
Number of zeroes placed after decimal will be $2.$
Antilog $-2.0258=0.009423$
Hence, B is the correct option.

Multiple choice logarithm and its uses basic mathematical concepts physics

What is the value of $[\log _{10} (5\log _{10} 100)]^{2}$?

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

The value of $[\log _{10}(5\log _{10} 100)]^{2}$ is
$=[\log _{10}(5\log _{10}10^2)]^2$

$=[\log _{10}(10\log _{10}10)]^2$     .....As $\log a^m=m\log a$
$=[\log _{10}(10\times 1)]^2$     ....As $\log _aa=1$
$=[\log _{10}10]^2$
$=[1]^2=1$