7.158. It is known that $\beta=10^{\frac{1}{1-\lg \alpha}}$ and $\gamma=10^{\frac{1}{1-\lg \beta}}$. Find the dependence of $\alpha$ on $\gamma$.
## Solution.

$\lg \beta=\frac{1}{1-\lg \alpha} ; \lg \gamma=\frac{1}{1-\lg \beta}=\frac{1}{1-\frac{1}{1-\lg \alpha}}=\frac{1-\lg \alpha}{-\lg \alpha}=-\frac{1}{\lg \alpha}+1$;

$\frac{1}{\lg \alpha}=1-\lg \gamma ; \lg \alpha=\frac{1}{1-\lg \gamma} ; \alpha=10^{1 /(1-\lg \gamma)}$.

Answer: $\alpha=10^{1 /(1-\lg \gamma)}$.