Question

If $3^x=5^y=(75{)}^z,showthatz=\frac{xy}{2x+y}$

Answer 1

Let $3^x=5^y=(75{)}^z=k\therefore 3=k^{\frac{1}{x}},5=k^{\frac{1}{y}}and75=k^{\frac{1}{z}}\Rightarrow 3\times 5^2=k^{\frac{1}{z}}\Rightarrow k^{\frac{1}{x}}\times k^{\frac{2}{y}}=k^{\frac{1}{z}}\Rightarrow k^{\frac{1}{x}+\frac{2}{y}}=k^{\frac{1}{z}}\Rightarrow \frac{1}{x}+\frac{2}{y}=\frac{1}{z}\Rightarrow \frac{y+2x}{xy}=\frac{1}{z}\therefore z=\frac{xy}{2x+y}$