Question

If $(4.781{)}^x=(0.4781{)}^y={10}^z$ then find the value of $-\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.

• A. $1$
• B. $0$
• C. $2$
• D. $-1$

Answer 1

The correct option is B $0$

Let

${\left(4.781\right)}^x={\left(0.4731\right)}^y={10}^z=k$

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

$x\log \left(4.781\right)=\log k$

$x=\frac{\log k}{\log \left(4.781\right)}$

Similarly,

$y=\frac{\log k}{\log \left(0.4781\right)},z=\frac{\log k}{\log 10}$

We have,

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

$\Rightarrow -\frac{1}{\frac{\log k}{\log \left(4.781\right)}}+\frac{1}{\frac{\log k}{\log \left(0.4781\right)}}+\frac{1}{\frac{\log k}{\log 10}}$

$\Rightarrow -\frac{\log \left(4.781\right)}{\log k}+\frac{\log \left(0.4781\right)}{\log k}+\frac{\log 10}{\log k}$

$\Rightarrow \frac{-\log \left(4.781\right)+\log \left(0.4781\right)+\log 10}{\log k}$

$\Rightarrow \frac{-\log \left(4.781\right)+\log \left(4.781\right)}{\log k}$

$\Rightarrow 0$

Hence, this is the answer.