How do you solve $Log_b 64 = 6$ ?

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Answer 1 · 2016-03-07T08:46:25

$b^6=64->b^6 = 2^6->(b^6)^(1/6) = (2^6)^(1/6) ->b = 2$ Or
$b^6=64->root6(b^6) =root6(64)->b=2$

First, use definition of logarithm to change to exponential form then solve for b by taking the sixth root of both sides.

How do you solve Log_b 64 = 6Log_b 64 = 6?