How do you solve #log_3 x=2#?

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Answer 1 · 2016-10-09T06:51:41

#x=9#

We can write log problems two different ways:

#log_(a)b=c#

#a^c=b#

So we can take the original question:

#log_3x=2#

and rewrite it to:

#3^2=x=9#