7.54 It is known that $\beta=10^{\frac{1}{1-\lg \alpha}}$ and $\gamma=10^{\frac{1}{1-\operatorname{lg} \beta}}$. Find the dependence of $\alpha$ on $\gamma$.
7.54 By logarithmizing the equalities with the base, we get $\lg \beta=\frac{1}{1-\lg \alpha}, \lg \gamma=\frac{1}{1-\lg \beta}$. Therefore, $\lg \gamma=\frac{1}{1-\frac{1}{1-\lg \alpha}}=\frac{\lg \alpha-1}{\lg \alpha}$,

from which $\lg \alpha=\frac{1}{1-\lg \gamma}$. According to the definition of the logarithm, we have $\alpha=10^{\frac{1}{1-\lg \gamma}}$.

Answer: $\alpha=10^{\frac{1}{1-\lg \gamma}}$.