Question

I need to solve y=2^x using logarithmic differentiation.

Answer 1

\[(a^x)' = a^x(\ln~a)\]

Answer 2

Would it not be just:
\[y' = 2^x(\ln 2)\]
?

Answer 3

I got the answer using that method but need this method.
I would take log of both sides but I get stuck

Answer 4

heres where I lock up