how do you prove that? (see the image) how would you simplify the left side to show it equals the right?
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Hint: $\displaystyle \log_a b \times \log _b a = 1$, so $\displaystyle \frac 1 {\log _a b} = \log _b a$
Originally Posted by rgchel how do you prove that? (see the image) how would you simplify the left side to show it equals the right? Here is a second way of looking at it. Because $\log_b(x)=\dfrac{\log(x)}{\log(b)}$ then $(\log_3(x))^{-1}=\dfrac{\log(3)}{\log(x)}$ Now on the LHS we have a common denominator.
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