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

1 Answer
Daan J.
Sep 5, 2015

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.

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