How do you solve #12^(y-2) =20#?

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Answer 1 · 2016-08-23T22:25:31

#y = ln2880/ln12#

Take the natural logarithm of both sides.

#ln(12^(y - 2)) = ln20#

#(y - 2)ln12 = ln20#

#yln12 - 2ln12 = ln20#

#yln12 = ln20 + ln144#

#y = ln2880/ln12#

Hopefully this helps!