Question

How do you solve `log4^(2x) = y + 7`?

Answer 1

It turns out this equation is equivalent to `y=2log(4)x-7`.

Explanation:

the core of this problem is that for any real positive real number `a` and any real number `b` we have `log(a^b)=blog(a)`. Therefore we van rewrite:
`log(4^(2x))=y+7`
as
`2xlog(4)=y+7`
A quick rearrangement gives
`y=2log(4)x-7`.

If you want to, you could rewrite `2log(4)` to `log(4^2)=log(16)`, or since `4=2^2`, `2log(4)=4log(2)`, but since they are all the same number and you can't write out `log(4)` in decimal notation without rounding off anyway, how you write it down doesn't really matter.