Question

If $(5.2{)}^x=(0.52{)}^y=10000$,then $\frac{1}{x}-\frac{1}{y}=0$

Answer 1

Given, ${\left(5.2\right)}^x={\left(0.52\right)}^y=10000\Rightarrow {\left(5.2\right)}^x={\left(0.52\right)}^y={10}^4$

${\left(5.2\right)}^x={10}^4$

taking $\log$ in both sides we get,

$\Rightarrow \log {\left(5.2\right)}^x=\log \left({10}^4\right)\Rightarrow x\log \left(5.2\right)=4\log \left(10\right)\Rightarrow x=\frac{4\log \left(10\right)}{\log \left(5.2\right)}\Rightarrow \frac{1}{x}=\frac{\log \left(5.2\right)}{4}$

Similarly with the above process we can find,

$\frac{1}{y}=\frac{\log \left(0.52\right)}{4}$

So, $\frac{1}{x}-\frac{1}{y}=\frac{\log \left(5.2\right)}{4}-\frac{\log \left(0.52\right)}{4}=\frac{\log \left(\frac{5.2}{0.52}\right)}{4}=\frac{1}{4}$