Question

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

Answer 1

Let, $(2.381{)}^x=(0.2381{)}^y={10}^z=k$

$⟹2.381=k^{\frac{1}{x}},0.2381=k^{\frac{1}{y}},10=k^{\frac{1}{z}}$

Now,

$10=\frac{2.381}{0.2381}$

$k^{\frac{1}{z}}=\frac{k^{\frac{1}{x}}}{\frac{1}{y}}$

$k^{\frac{1}{z}}=k^{\frac{1}{x}-\frac{1}{y}}$ ------ Quotient law of exponents

$\frac{1}{z}=\frac{1}{x}-\frac{1}{y}$ ------ Base is same, powers can be equated

$\therefore \frac{1}{z}-\frac{1}{x}+\frac{1}{y}=0$