Sum of three weighted logarithms with three different bases is equal to an integer

Question

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 be

Source: 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.

Answer 1

\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, $a=27$.

Answer 2

Here's a possible method. Firstly, it has been given that $a,b,c>1$, which means that, if: $$x=2\log a$$ $$y=3\log b$$ $$z=5\log c$$ Where all the bases are $10$ (for convenience, any other base is possible), then we have: $x,y,z>0$. Now, using $\log_p q=\frac {\log q}{\log p}$, our given condition reduces to: $$\frac xy+\frac yz+\frac zx=3$$ Since $x,y,z>0$, AM-GM applies, so $\frac xy=\frac yz=\frac zx$, since minima is reached. This means that each of the quantities are equal to $1$, so $x=y=z$. So, $2\log a=3\log 9$ which means that $a^2=3^6$ which means $a=27$.

Edit: Since I see that AM-GM has been used, another way to obtain the equality would be to see that: $$\left(\sqrt[3] {\frac xy}\right)^3+\left(\sqrt[3]{\frac yz}\right)^3+\left(\sqrt[3]{\frac zx}\right)^3=3\left(\sqrt[3]\frac xy\right)\left(\sqrt[3] {\frac yz}\right)\left(\sqrt[3]{\frac zx}\right)$$ So, since $p^3+q^3+r^3=3pqr$ implies $p+q+r=0$ or $p=q=r$, the former of which is impossible, we are forced to conclude that $x=y=z$.