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# CAT Questions on Logarithms:

Logarithms is one of the basic topic of quantitative aptitude. You can learn logarithms rules useful for CAT exam. We have provided some practice questions on logarithms for CAT. We hope this log questions and answers will be useful to understand and remember concepts related to logarithms.

Question 1: $\log_{3}{81\sqrt3}+\log_{5\sqrt{5}}{625}-\log_{7}{49\sqrt{7}}$ = ?

a) 5
b) 6
c) 7
d) None of these

Question 2: If $log_8^{log_3^{log_2^{2+3\sqrt{x}}}} = \frac{1}{3}$. Find x.

a) 512
b) 289
c) 170
d) None of these

Question 3: Let $f(n)=n^{log_{24}37}$. Find the remainder when $f(2)\times f(3) \times f(4)^{2}\times f(6)$ is divided by 13.

a) 4
b) 7
c) 1
d) None of these

Question 4: If $\log_{125} 729$ = x, then what is the value of $\log_{81} 5625$?

a) $\frac{4+x}{2x}$
b) $\frac{6+2x}{3x}$
c) $\frac{4+2x}{3x}$
d) $\frac{6+x}{2x}$

Question 5:  Find the value of $log_{10} 10 + log_{10}10^{2} + …….. + log_{10}10^{n}$

a) $n*(2n+1)$
b) $\frac{n*(n+1)}{2}$
c) $\frac{3n*(n+1)}{2}$
d) $\frac{n*(n^{2}+1)}{2}$

Solutions for CAT Questions on Logarithms:

$\log_{3}{81\sqrt3}+\log_{5\sqrt{5}}{625}-\log_{7}{49\sqrt{7}}$
$=\frac{9}{2}+\frac{8}{3}-\frac{5}{2}$
$=\frac{14}{3}$

$log_8^{log_3^{log_2^{2+3\sqrt{x}}}} = \frac{1}{3}$
==>${log_3^{log_2^{2+3\sqrt{x}}}} = 2$
==>${log_2^{2+3\sqrt{x}}} = 9$
==> $2+3\sqrt{x} = 2^9 = 512$
==> $\sqrt{x} = 170$ ==> x=28900
Since there is no such option, the correct option to choose is D – None of these.

$f(2)\times f(3) \times f(4)^{2}\times f(6)$
= $2^{log_{24}37}\times 3^{log_{24}37}\times 16^{log_{24}37}\times 6^{log_{24}37}$
=$576^{log_{24}37}=24^{2{log_{24}37}}=24^{log_{24}37^{2}} = 37^{2}$
We have to find the remainder when $37^{2}$ is divided by 13.
$37^{2} =(39-2)^{2}$
Thus, the remainder will be $(-2)^{2}=4$

$\log_{125} 729$ = $\log_{5} 9$ = x
=> $\log_{9} 5 = \frac{1}{x}$
=> $\log_{3} 5 = \frac{2}{x}$
$\log_{81} 5625 = \log_{81} 625 + \log_{81} 9 = \log_{3} 5 + \frac{1}{2}$
= $\frac{2}{x} + \frac{1}{2} = \frac{4+x}{2x}$.

$log_{10} 10 + log_{10}10^{2} + …….. + log_{10}10^{n}$
=> $log_{10} 10 + 2*log_{10}10 + …….. + n*log_{10}10$
=> $log_{10}10{1 + 2 + 3 . . . . . .n}$
=> $log_{10}10{\frac{n*(n+1)}{2}}$
=> $\frac{n*(n+1)}{2}$