How do you solve #log_3x=9#?

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How do you solve #log_3x=9#?

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Answer 1 · 2016-10-24T07:11:04

#x=3^9#

Explanation:

When I write #log_(a)b=c#, I ask to what power I raise the base #a# tp get #b#; here #a^c=b#. By way of example, #log_(10)100=2,# i.e. #10^2=100, 10^3=1000 etc.#

Given that #log_(3)x=9#, then #x=3^9#, whatever this is it is large!

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How do you solve #log_3x=9#?

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