If $4^{A} + 9^{B} = 10^{c},$ where $A = {log}_{16} 4, B = {log}_{3} 9$ & $C = {log}_{x} 83$ , then the value of $x$ is

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Solution

$A = {log}_{16} 4 = {log}_{4^{2}} 4 = \frac{1}{a} {log}_{4} 4 = \frac{1}{2} . . . . [∵ {log}_{a^{n}} b^{m} = \frac{m}{n} {log}_{a} b, {log}_{a} a = 1] B = {log}_{3} 9 = {log}_{3} 3^{2} = 2 {log}_{3} 3 = 2$
Therefore,
$4^{A} + 9^{B} = 10^{C} \Rightarrow 4^{\frac{1}{2}} + 9^{2} = 10^{C} \Rightarrow 2 + 81 = 10^{C} \Rightarrow 83 = 10^{C} \Rightarrow {log}_{10} 83 = C = {log}_{x} 83 \Rightarrow x = 10$

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