How do you solve #log_9x = 1.5#?

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How do you solve #log_9x = 1.5#?

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Answer 1 · 2016-06-11T12:23:50

I found #x=27#

Explanation:

We can write it trying to manipualte #1.5# into #3/2#:
#log_9(x)=3/2#
we use the definition of log:
#x=9^(3/2)#
and the property of exponents:
#x=sqrt(9^3)#
#x=27#

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How do you solve #log_9x = 1.5#?

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