How do you solve #2^x=27#?

1 Answer
Kevin B.
Nov 13, 2016

The answer is #log_2(27)#

Explanation:

To solve this problem we need the inverse operation of exponents: logarithms. If #2# raised to some power #x# is equal to #27#, then the logarithm of base #2# and an argument of #27# is equal to #x#.

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