Question

If $3^x=5^y={75}^z$ show that $z=\frac{xy}{2x+y}$

Answer 1

Solve for the required value

Given that $3^x=5^y={75}^z$

Let $3^x=5^y={75}^z=k$

$\Rightarrow 3=k^{\frac{1}{x}},5=k^{\frac{1}{y}},75=k^{\frac{1}{z}}$

$\Rightarrow 5^2\times 3=k^{\frac{1}{z}}$ $\left[\because 75=25\times 3\right]$

$\Rightarrow {\left(k^{\frac{1}{y}}\right)}^2\times \left(k^{\frac{1}{x}}\right)=\left(k^{\frac{1}{z}}\right)$

$\Rightarrow$ $k^{\left(\frac{2}{y}+\frac{1}{x}\right)}=k^{\frac{1}{z}}\left[\because a^m\times a^n=a^{\left(m+n\right)}\right]$

Now comparing the exponents of $k$

$$\begin{gathered} \Rightarrow \frac{2}{y}+\frac{1}{x}=\frac{1}{z} \\ \Rightarrow \frac{2x+y}{xy}=\frac{1}{z} \end{gathered}$$

$\Rightarrow$ $z=\frac{xy}{2x+y}$

Hence it is proved that $z=\frac{xy}{2x+y}$