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So I've been trying to solve a question based on logarithms and it says this:-

Let $a$, $b$ and $c$ be real numbers, each greater than 1, such that $$\frac{2}{3}\log_{b}{a} + \frac{3}{5}\log_{c}{b} + \frac{5}{2}\log_{a}{c} = 3.$$ If $b = 9$, then what is the value of $a$?

So far, after trying everything using elementary logarithmic operations, I arrived here:-

$$\log_{b^3}{a^2} + \log_{c^5}{b^3} + \log_{a^2}{c^5} = 3$$

Now, I'm proposing that all of those logarithms to be equal to 1. But I'm not sure if that proposition holds true everywhere. I have tried to assume a real number $k > 1$ where $\log c^5 = k\log a^2$, and then try to see if that holds true for the relation given, but the expression which yields after that is not something pleasant.

Any help would be appreciated.

rikusp2002
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Surely its coming out to be A.M. equal to G.M. that means all three no.s are equal so you can form three equations and find the answer.

  • Understood, so I have to use the base change law to get an equation like $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 3$$, which could be used to get the relation of AM-GM. – rikusp2002 Jun 05 '21 at 03:36
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    Yes, you got it right –  Jun 05 '21 at 04:36