# Logarithms Questions for IIFT PDF

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## Logarithms Questions for IIFT PDF

Download important IIFT Logarithms Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice Logarithms questions and answers for IIFT exam.

Practice IIFT Mock Tests

Question 1: If $log_3 2, log_3 (2^x – 5), log_3 (2^x – 7/2)$ are in arithmetic progression, then the value of x is equal to

a) 5

b) 4

c) 2

d) 3

Question 2: If $log_y x = (a*log_z y) = (b*log_x z) = ab$, then which of the following pairs of values for (a, b) is not possible?

a) (-2, 1/2)

b) (1,1)

c) (0.4, 2.5)

d) ($\pi$, 1/ $\pi$)

e) (2,2)

Question 3: If $f(x) = \log \frac{(1+x)}{(1-x)}$, then f(x) + f(y) is

a) $f(x+y)$

b) $f{\frac{(x+y)}{(1+xy)}}$

c) $(x+y)f{\frac{1}{(1+xy)}}$

d) $\frac{f(x)+f(y)}{(1+xy)}$

Question 4: If $\log_{2}{x}.\log_{\frac{x}{64}}{2}=\log_{\frac{x}{16}}{2}$. Then x is

a) 2

b) 4

c) 16

d) 12

Question 5: Find the value of x from the following equation:
$\log_{10}{3}+\log_{10}(4x+1)=\log_{10}(x+1)+1$

a) 2/7

b) 7/2

c) 9/2

d) None of the above

Question 6: If $log_{10} x – log_{10} \sqrt[3]{x} = 6log_{x}10$ then the value of x is

a) 10

b) 30

c) 100

d) 1000

Question 7: $(1+5)\log_{e}3+\frac{(1+5^{2})}{2!}(\log_{e}3)^{2}+\frac{(1+5^{3})}{3!}(\log_{e}3)^{3}+…$

a) 12

b) 244

c) 243

d) 245

Question 8: If $log(2^{a}\times3^{b}\times5^{c} )$is the arithmetic mean of $log ( 2^{2}\times3^{3}\times5)$, $log(2^{6}\times3\times5^{7} )$, and $log(2 \times3^{2}\times5^{4} )$, then a equals

Question 9: If $\log_{2}({5+\log_{3}{a}})=3$ and $\log_{5}({4a+12+\log_{2}{b}})=3$, then a + b is equal to

a) 59

b) 40

c) 32

d) 67

Question 10: $\frac{1}{log_{2}100}-\frac{1}{log_{4}100}+\frac{1}{log_{5}100}-\frac{1}{log_{10}100}+\frac{1}{log_{20}100}-\frac{1}{log_{25}100}+\frac{1}{log_{50}100}$=?

a) $\frac{1}{2}$

b) 10

c) 0

d) −4

$2 log (2^x – 5) = log 2 + log (2^x – 7/2)$
Let $2^x = t$
=> $(t-5)^2 = 2(t-7/2)$
=> $t^2 + 25 – 10t = 2t – 7$
=> $t^2 – 12t + 32 = 0$
=> t = 8, 4
Therefore, x = 2 or 3, but $2^x$ > 5, so x = 3

$log_y x = ab$
$a*log_z y = ab$ => $log_z y = b$
$b*log_x z = ab$ => $log_x z = a$
$log_y x$ = $log_z y * log_x z$ => $log x/log y$ = $log y/log z * log z/log x$
=> $\frac{log x}{log y} = \frac{log y}{log x}$
=> $(log x)^2 = (log y)^2$
=> $log x = log y$ or $log x = -log y$
So, x = y or x = 1/y
So, ab = 1 or -1
Option 5) is not possible

If $f(x) = \log \frac{(1+x)}{(1-x)}$ then $f(y) = \log \frac{(1+y)}{(1-y)}$

Also Log (A*B)= Log A + Log B

f(x)+f(y) = $\log \frac{(1+x)(1+y)}{(1-x)(1-y)}$ solving we get $\log { \frac{1+ \frac{(x+y)}{(1+xy)}}{1- \frac{(x+y)}{(1+xy)}}}$

Hence option B.

$\log_{2}{x}.\log_{\frac{x}{64}}{2}=\log_{\frac{x}{16}}{2}$

i.e. $\frac{log{x}}{log{2}} * \frac{log_{2}}{log{x}-log{64}} = \frac{log{2}}{log{x}-log{16}}$

i.e. $\frac{log{x} * (log{x}-log{16})}{log{x}-log{64}}$ = $\log{2}$

let t = log x

Therefore,  $\frac{t * (t-log{16})}{t-log{64}}$ = $\log{2}$

$t^2-4*log 2*t = t*log 2-6*(log 2)^2$

I.e. $t^2-5*log 2*t-6*(log 2)^2$ = 0

I.e. $t^2-3*log 2*t-2*log 2*t-6*(log 2)^2$ = 0

i.e. $t*(t-3*log 2)-2*log 2*(t-3*log 2)$ = 0

i.e $t=2*log 2$ or $t=3*log 2$

i.e $log x=log 4$ or $log x=log 8$

therefore $x=4$ or $8$

therefore our answer is option ‘B’

$\log_{10}{3}+\log_{10}(4x+1)=\log_{10}(x+1)+1$ can be written as

$\log_{10}{3}+\log_{10}(4x+1)=\log_{10}(x+1)+\log_{10}{10}$

We know that $\log_{10}{a}+\log_{10}{b}=\log_{10}{ab}$

$\log_{10}{3*(4x+1)}=\log_{10}{(x+1)*10}$

$12x+3=10x+10$

$x=7/2$. Hence, option B is the correct answer.

$\log_{10} x – \log_{10} \sqrt[3]{x} = 6\log_{x}10$
Thus, $\dfrac{\log {x}}{\log {10}}$ – $\dfrac{1}{3}*\dfrac{\log {x}}{\log {10}}$ = $6*\dfrac{\log{10}}{\log{x}}$
=> $\dfrac{2}{3}*\dfrac{\log {x}}{\log {10}}$ = $6*\dfrac{\log{10}}{\log{x}}$
Thus, => $\dfrac{1}{9}*(\log{x})^2 = (\log{10})^2$
Thus, $x = 1000$
Hence, option D is the correct answer.

Splitting the above mentioned series into two series

A = $\log_{e}3+\frac{1}{2!}(\log_{e}3)^{2}+\frac{1}{3!}(\log_{e}3)^{3}+…$

B = $5\log_{e}3+\frac{5^{2}}{2!}(\log_{e}3)^{2}+\frac{5^{3}}{3!}(\log_{e}3)^{3}+…$

We know that $e^{x}$ =$1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+…$

So  $e^{x}-1$ = $x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+…$

On solving two series A and B

A = $\log_{e}3+\frac{1}{2!}(\log_{e}3)^{2}+\frac{1}{3!}(\log_{e}3)^{3}+…$ =$e^{\log_{e}3}-1$ = $3-1$ =$2$

B = $5\log_{e}3+\frac{5^{2}}{2!}(\log_{e}3)^{2}+\frac{5^{3}}{3!}(\log_{e}3)^{3}+…$=$e^{\log_{e}3^{5}}-1$=$3^{5}-1$=$242$

A+B = $2 + 242$ = $244$

$log(2^{a}\times3^{b}\times5^{c} )$ = $\frac{log ( 2^{2}\times3^{3}\times5) + log(2^{6}\times3\times5^{7} ) + log(2 \times3^{2}\times5^{4} ) }{3}$

$log(2^{a}\times3^{b}\times5^{c} )$ = $\frac{log ( 2^{2+6+1}\times3^{3+1+2}\times5^{1+7+4}) }{3}$

$log(2^{a}\times3^{b}\times5^{c} )$ = $\frac{log ( 2^{9}\times3^{6}\times5^{12}) }{3}$

$3log(2^{a}\times3^{b}\times5^{c} )$ = $log ( 2^{9}\times3^{6}\times5^{12})$
Hence, 3a = 9 or a = 3

$\log_{2}({5+\log_{3}{a}})=3$
=>$5 + \log_{3}{a}$ = 8
=>$\log_{3}{a}$ = 3
or $a$ = 27

$\log_{5}({4a+12+\log_{2}{b}})=3$
=>$4a+12+\log_{2}{b}$ = 125
Putting $a$ = 27, we get
$\log_{2}{b}$ = 5
or, $b$ = 32

So, $a + b$ = 27 + 32 = 59
Hence, option A is the correct answer.

We know that $\dfrac{1}{log_{a}{b}}$ = $\dfrac{log_{x}{a}}{log_{x}{b}}$

Therefore, we can say that $\dfrac{1}{log_{2}{100}}$ = $\dfrac{log_{10}{2}}{log_{10}{100}}$

$\Rightarrow$ $\frac{1}{log_{2}100}-\frac{1}{log_{4}100}+\frac{1}{log_{5}100}-\frac{1}{log_{10}100}+\frac{1}{log_{20}100}-\frac{1}{log_{25}100}+\frac{1}{log_{50}100}$

$\Rightarrow$ $\dfrac{log_{10}{2}}{log_{10}{100}}$-$\dfrac{log_{10}{4}}{log_{10}{100}}$+$\dfrac{log_{10}{5}}{log_{10}{100}}$-$\dfrac{log_{10}{10}}{log_{10}{100}}$+$\dfrac{log_{10}{20}}{log_{10}{100}}$-$\dfrac{log_{10}{25}}{log_{10}{100}}$+$\dfrac{log_{10}{50}}{log_{10}{100}}$

We know that $log_{10}{100}=2$

$\Rightarrow$ $\dfrac{1}{2}*[log_{10}{2}-log_{10}{4}+log_{10}{5}-log_{10}{10}+log_{10}{20}-log_{10}{25}+log_{10}{50}]$

$\Rightarrow$ $\dfrac{1}{2}*[log_{10}{\dfrac{2*5*20*50}{4*10*25}}]$

$\Rightarrow$ $\dfrac{1}{2}*[log_{10}10]$

$\Rightarrow$ $\dfrac{1}{2}$

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