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.

Related questions
See all questions in Logarithmic Models
Impact of this question
1736 views around the world
You can reuse this answer
Creative Commons License