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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