Let $a, b$ and $c$ be real numbers, each greater than $1$, such that
$$\frac2
3
\log_b (a) +
\frac3
5
\log_c
(b) +
\frac5
2
\log_a
(c) = 3$$If the value of $b$ is $9$, then the value of $a$ must beSource: ISI BMath UGA 2017
I'm not very familiar on how to solve questions relating to logarithms of the above sort, but I started with the base change rule: $ \log_c (b) = \frac{1}{\log_b(c)}$, $ \log_a (c) = \frac{\log_b (c)}{\log_b(a)}$, $b=9$, plugging the information in:
$$ \frac23 \log_9 (a) + \frac35 \frac{1}{\log_9(c)} + \frac52 \frac{\log_9(c)}{\log_9(a)}=3 \tag{1}$$
Making the denominator as one:
$$ \frac23 \log_9(a)^2 \log_9(c) + \frac35 \log_9 (a) + \frac52 \log_9 (a) \log_9(c)^2 = \log_9(c) \log_9(a)$$
This doesn't seem the way to go.
Observation: The product of coefficient in the lhs of equation-(1) is just one.
I thought maybe if I cube both sides then there will be a term of one on the LHS .. but the rest of the expressions is quite ugly, so it doesn't seem to be of much help.
Best Answer
\begin{align*} \frac23\frac{\log a}{\log b}+\frac35\frac{\log b}{\log c}+\frac52\frac{\log c}{\log a}\ge3\left(\frac23\frac{\log a}{\log b}\cdot\frac35\frac{\log b}{\log c}\cdot\frac52 \frac{\log c}{\log a}\right)^\frac13=3 \end{align*}
with the equality holds if and only if $\displaystyle \frac23\frac{\log a}{\log b}=\frac35\frac{\log b}{\log c}=\frac52\frac{\log c}{\log a}$
So, $a=27$.