If $3(4^h)=4(2^k)$ and $9(8^h)=20(4^k)$,show that $2^h = \frac{4}{5}$

Question

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?

Answer 1

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 $(2^h)^3=64/45$.

Answer 2

André Nicolas gave a very simple solution.

If you really want to use logarithms, then $$\log(3)+h\log(4)=\log(4)+k \log(2)$$ $$\log(9)+h\log(8)=\log(20)+k \log(4)$$ Now, taking into account $9=3^2$, $8=2^3$, $4=2^2$, the expressions write $$\log(3)+2h\log(2)=2\log(2)+k \log(2)$$ $$2\log(3)+3h\log(2)=\log(20)+2k\log(2)$$ and you have two linear equations in $h,k$ easy to solve (for example, multiply the first by $2$ and subtract from the second to get $h$).

I am sure that you can take from here.

Answer 3

${\left( {3({4^h}) = 4({2^k})} \right)^2 \rightarrow 9({16^h}) = 16({4^k}) \choose 9({8^h}) = 20({4^k})}$$\Rightarrow$ $\frac{{9({{16}^h})}}{{9({8^h})}} = \frac{{16({4^k})}}{{20({4^k})}} \Rightarrow {2^h} = \frac{4}{5}$