How do you solve #log_b 9=2#?

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Answer 1 · 2015-12-10T05:46:45

#b= 3#
Change to the exponential form as explained below.

Given #log_b9= 2#

Change this equation to its exponential form, since

#log_ax= y iff a^y = x#

#log_b9=2#

#b^2= 9#
#b^ 2= 3^2#
#b= 3#

Remember, if the exponents are the same, then the answer is the base.