Q. If $(4)^{\log_9 3} + (9)^{\log_2 4} = (10)^{\log_x 83}$, then X is equal to

COMEDKCOMEDK 2012Probability - Part 2 Report Error

A

10

79%

B

4

13%

C

-10

4%

D

-4

4%

Solution:

$\left(4\right)^{\log _{_{3^2^{^3}}}} + \left(9\right)^{\log _{_{2^{2^2}}}} = \left(10\right)^{\log _{x}}^{ 83}$
$\Rightarrow \left(4\right)^{12} + 9^{2} = \left(10\right)^{\log _{x} 83} $
[Using $ x 0\log_a y = y \, log_a x$]
$\Rightarrow \left(83\right)^{1} = \left(83\right)^{\log _{x} 10} \Rightarrow 1 =\log_{x} 10 \Rightarrow x = 10 $

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