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

1 Answer
Nov 13, 2016

Answer:

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#.