Alg3.3.12

If $a^x=b^y=c^z\,$ and $\frac{b}{a}=\frac{c}{b}\,$ ,show that $\frac{y}{x}=\frac{2z}{x+z}\,$

Let $a^x=b^y=c^z=k\,$

$a=k^{\frac{1}{x}},b=k^{\frac{1}{y}},c=k^{\frac{1}{z}}\,$

Substituting these values in the given condition $b^2=ac\,$

we get

$(k^{\frac{1}{y}})^2=(k^{\frac{1}{x}}\cdot (k^{\frac{1}{z}})\,$

$k^{\frac{2}{y}}=k^{\frac{1}{x}+\frac{1}{z}}\,$

Since bases are equal,

$\frac{2}{y}=\frac{1}{x}+\frac{1}{z}\,$

Simplifying

$2zx=y(z+x)\,$

Dividing by x on both sides we get

$2z=\frac{y}{x}(z+x),\frac{y}{x}=\frac{2z}{z+x}\,$