If $3(4^h)=4(2^k)$ and $9(8^h)=20(4^k)$,show that $2^h = \frac{4}{5}$ If $3(4^h)=4(2^k)$ and $9(8^h)=20(4^k)$,show that $2^h = \frac{4}{5}$.
I tried to substitute the equation 1 into equation 2 so that I can find the value of $k$ or $h$, but it did not work as the base is not the same (I cannot compare to find the answer)
How can I find the answer?

We can see that $2^h4^{h-k}=\frac{20}{9}$. On the other hand $2^{2h-k}=\frac{4}{3}$. So $\frac{4^{h-k}}{2^{k-h}}=\frac{5}{3}$. This implies $2^{3h-3k}=\frac{5}{3}$. Since $2^k=\frac{3}{4}2^{2h}$ we have $2^{3k}=\frac{27}{64}2^{6h}$. Therefore, $\frac{2^{3h}}{2^{3k}}=\frac{64}{27}\frac{2^{3h}}{2^{6h}}=\frac{64}{27}2^{-3h}=\frac{5}{3}$. This implies $(2^h)^3=64/45$.
