NIMCET 2016 — Mathematics PYQ

Concept:
Logarithm Rule:
$\log a^x=x\log a$

**Calculations:**

Given, $3^x=4^{x-1}$  

Taking logarithm on both sides, we get  
$\log 3^x=\log 4^{x-1}$  

$\Rightarrow x\log 3=(x-1)\log 4$  

$\Rightarrow x\log 3=x\log 4-\log 4$  

$\Rightarrow x(\log 4-\log 3)=\log 4$  

$\Rightarrow x=\frac{\log 4}{\log 4-\log 3}$  

$\Rightarrow x=\frac{\log 2^2}{\log 2^2-\log 3}$  

$\Rightarrow x=\frac{2\log 2}{2\log 2-\log 3}$  

$\Rightarrow x=\frac{2\log 2}{2{\log }_{3}2-1}$  

Hence, If $3^x=4^{x-1}$, then  
$x=\frac{2\log 2}{2{\log }_{3}2-1}$