I came across a problem; $a^x=b^y=c^z$ and $b^2=ac$. It is required to show $\frac{1}{x}+\frac{1}{z}=\frac{2}{y}$. I have tried the following steps-
\begin{equation*} b^2=ac \\ b=\sqrt{ac} \\ b=a^{1/2}\ast c^{1/2} \end{equation*} So \begin{equation*} a^x=a^{1/2}\ast c^{1/2} \\ c^z=a^{1/2}\ast c^{1/2} \end{equation*}
After that I tried to find the value of $x$ and $z$ but I could not solve it this way. Can anyone please show me how I can solve this?
On
Taking logarithms, we get $$ x\log a = y \log b = z \log c $$ and $$ 2 \log b = \log a + \log c $$ Hence \begin{align*} \frac 2y &= \frac{2\log b}{y \log b}\\ &= \frac{\log a}{y\log b} + \frac{\log c}{y \log b}\\ &= \frac{\log a}{x \log a} + \frac{\log c}{z \log c}\\ &= \frac 1x + \frac 1z \end{align*}
Let $a^x=b^y=c^z=k\implies a=k^{1/x}$ etc.
$$\displaystyle b^2=ca\implies(k^{1/y})^2=k^{1/x}\cdot k^{1/z}$$
$$\displaystyle\implies k^{\frac2y}=k^{\frac1x+\frac1z} $$
Assuming $k\ne0,1,$
$$ \dfrac2y=\dfrac1x+\dfrac1z$$