How would I solve : $\log_a(b)=log_b(a)\}$ such that a and b is not equal to 1 and a is not equal to b. Would Diophantine equation be the way to go for this equation?
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Note that $$\log_a b=\frac{1}{\log_b a}$$ so the question is equivalent to $$(\log_a b)^2=1$$ so $a=b$ or $a=b^{-1}$.
Rene Schipperus
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While trying to solve the question I also found a = b by applying the properties of log. I followed and understand each of your steps, but I don't see why a = b^-1. I understand the a = b. – coderhk Dec 07 '17 at 03:12
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1What if $\log_a b=-1$? – Rene Schipperus Dec 07 '17 at 03:15
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Recall $\log_a(b)=\frac{\ln b}{\ln a}$. Your equation can be rewritten as $\frac{\ln b}{\ln a}=\frac{\ln a}{\ln b}$. You should be able to solve this now. Since this looks like a homework problem, I will not reveal more details.
edm
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