# Logarithm Questions for XAT Set-2 PDF

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## Logarithm Questions for XAT Set-2 PDF

Download important Logarithm Questions Set-2 for XAT  Set-2 PDF based on previously asked questions in the XAT exam. Practice Logarithm Questions  Set-2 PDF for the XAT exam.

Question 1: Let x and y be positive real numbers such that
$\log_{5}{(x + y)} + \log_{5}{(x – y)} = 3,$ and $\log_{2}{y} – \log_{2}{x} = 1 – \log_{2}{3}$. Then $xy$ equals

a) 150

b) 25

c) 100

d) 250

Question 2: Sham is trying to solve the expression:
$\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + …….. + \log \tan 89^\circ$.

a) 1

b) $\frac{1}{\sqrt{2}}$

c) 0

d) -1

Question 3: If $\log_{10}{11} = a$ then $\log_{10}{\left(\frac{1}{110}\right)}$ is equal to

a) $-a$

b) $(1 + a)^{-1}$

c) $\frac{1}{10 a}$

d) $-(a + 1)$

Question 4: Find the value of $\log_{10}{10} + \log_{10}{10^2} + ….. + \log_{10}{10^n}$

a) $n^{2} + 1$

b) $n^{2} – 1$

c) $\frac{(n^{2} + n)}{2}.\frac{n(n + 1)}{3}$

d) $\frac{(n^{2} + n)}{2}$

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

a) $\frac{1}{2}$

b) 10

c) 0

d) −4

Question 6: 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 7: 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 8: If x is a real number such that $\log_{3}5= \log_{5}(2 + x)$, then which of the following is true?

a) 0 < x < 3

b) 23 < x < 30

c) x > 30

d) 3 < x < 23

Question 9: $(1+5)\log_{e}3+\frac{(1+5^{2})}{2!}(\log_{e}3)^{2}+\frac{(1+5^{3})}{3!}(\log_{e}3)^{3}+…$

a) 12

b) 244

c) 243

d) 245

Question 10: 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

We have, $\log_{5}{(x + y)} + \log_{5}{(x – y)} = 3$

=> $x^2-y^2=125$……(1)

$\log_{2}{y} – \log_{2}{x} = 1 – \log_{2}{3}$

=>$\ \frac{\ y}{x}$ = $\ \frac{\ 2}{3}$

=> 2x=3y   => x=$\ \frac{\ 3y}{2}$

On substituting the value of x in 1, we get

$\ \frac{\ 5x^2}{4}$=125

=>y=10, x=15

Hence xy=150

$\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + …….. + \log \tan 89^\circ$.

=$\log \tan 1^\circ + \log \tan 89^\circ + \log \tan 2^\circ + \log \tan 88^\circ …….. + \log \tan 45^\circ$.

=$\log\ \left(\tan\ 1^0\cdot\tan\ 89^0\right)\times\log\ \left(\tan\ 2^0\cdot\tan\ 88^0\right)\ ………………………\log\ \left(\tan\ 45^0\right)$

tan $45^0$ = 1

$\log\ \left(\tan\ 45^0\right)\ =\ 0$

$\therefore$ $\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + …….. + \log \tan 89^\circ$ = 0

$\log_{10}{\left(\frac{1}{110}\right)}$

$\log_a\left(\ \frac{\ x}{y}\right)\ =\ \log_ax-\log_ay$

$\log_{10}{\left(\frac{1}{110}\right)}$ = $=\ \log_{10}1-\log_{10}110$

= 0$-\log_{10}110$

=$-\log_{10}11\times\ 10$

=$-\left(\log_{10}11+\log_{10}10\right)$

= -(a+1)

$\log_{10}{10} + \log_{10}{10^2} + ….. + \log_{10}{10^n}$

Since $\log_aa\$ = 1

$\log_{10}{10} + \log_{10}{10^2} + ….. + \log_{10}{10^n}$ = 1+2+….n

=$\ \frac{\ n\left(n+1\right)}{2}$

=$\frac{(n^{2} + n)}{2}$

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 $log_{10}{100}=2$

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

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

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

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

$\log_{2}({5+\log_{3}{a}})=3$
=>$5 + \log_{3}{a}$ = 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.

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

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

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

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

$1 < \log_{3}5 < 2$
=> $1 < \log_{5}(2 + x) < 2$
=> $5 < 2 + x < 25$
=> $3 < x < 23$

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 = $2 + 242$ = $244$

$\log_{10} x – \log_{10} \sqrt[3]{x} = 6\log_{x}10$
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, => $\dfrac{1}{9}*(\log{x})^2 = (\log{10})^2=1$
Thus, $(\log{x})^2 = 9$
Thus $\log x = 3$ or $-3$
Thus, $x = 1000$ or $\dfrac{1}{1000}$