Question

How do you solve `8^(x-y)=3^x` and `8^y=2^(x-3)`?

Answer 1

`x = log(8)/(log(3/4)), y = (log(8)/(log(3/4))-3)/3`

Explanation:

From `8^y =2^{x-3}` knowing that `8 = 2^3` we can write
`2^{3y} = 2^{x-3}` so we conclude that `3y=x-3`. Operating now on `8^{x-y}=3^x->2^{3(x-y)}=3^x` but `3y = x-3` substituting we get
`2^{2x+3}=3^x->8 times 4^x= 3^x->(3/4)^x=8`. Finally `x = log(8)/(log(3/4))`. Solving for `y` we get `y = (log(8)/(log(3/4))-3)/3`