How do you evaluate #g(x)=log_ax# for #x=a^2#?

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Answer 1 · 2018-05-31T14:36:03

#g(x)=2#

Inputting #a^2# for #x#, we get

#g(x)=log_a(a^2)#

At this point, the key realization here is that we can leverage the logarithm property

#log_a(a)=1#

#log_a# and #a# cancel each other out, so we essentially have

#g(x)=cancel(log_a)(cancela^2)#

#=>g(x)=2#

Hope this helps!