Question

If $(2.381{)}^x=(0.2381{)}^y={10}^z$ then find the value of $\frac{-1}{x}+\frac{1}{y}+\frac{1}{z}$.

Answer 1

Let

${\left(2.381\right)}^x={\left(0.2831\right)}^y={10}^z=k$

$\therefore {\left(2.381\right)}^x=k$

$x\log \left(2.381\right)=\log k$

$x=\frac{\log k}{\log \left(2.381\right)}$

Similarly,

$y=\frac{\log k}{\log \left(0.2381\right)},z=\frac{\log k}{\log 10}$

We have,

$\Rightarrow -\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

$\Rightarrow -\frac{1}{\frac{\log k}{\log \left(2.381\right)}}+\frac{1}{\frac{\log k}{\log \left(0.2381\right)}}+\frac{1}{\frac{\log k}{\log 10}}$

$\Rightarrow -\frac{\log \left(2.381\right)}{\log k}+\frac{\log \left(0.2381\right)}{\log k}+\frac{\log 10}{\log k}$

$\Rightarrow \frac{-\log \left(2.381\right)+\log \left(0.2381\right)+\log 10}{\log k}$

$\Rightarrow \frac{-\log \left(2.381\right)+\log \left(2.381\right)}{\log k}$

$\Rightarrow 0$

Hence, this is the answer.