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How do you solve #4(3^x)=20#?

1 Answer

Jan 15, 2017

#x=log_3 5#

Explanation:

First you would divide all the terms by 4:

#3^x=5#

then, through the equivalence:

#b^x=ahArrx=log_ba#,

you get

#x=log_3 5#

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