$a$,$b$ and $c$ are real numbers each bigger than 1 such that
$$\frac{2}{3}log_b{a} + \frac{3}{5}log_c{b} + \frac{5}{2}log_a{c} = 3$$
If the value of $b$ is 9 what must be the value of $a$?
I tried putting in the value of $b$ as 9 and then used the base change formula but couldn’t come up with a solution. Can someone help me out with the solution?
$log_b a = \frac {\log a}{\log b}$
$\frac {2\log a}{3\log b} + \frac {3\log b}{5\log c} + \frac {5\log c}{2\log a} = 3$
let $\alpha, \beta, \gamma = 2\log a, 3\log b, 5\log c$ respectively
The AM-GM inequality.
$\frac 13 (\frac {\alpha}{\beta} + \frac {\beta}{\gamma} + \frac {\gamma}{\alpha}) \ge (\frac {\alpha}{\beta} \frac {\beta}{\gamma} \frac {\gamma}{\alpha})^\frac 13$
$(\frac {\alpha}{\beta} \frac {\beta}{\gamma} \frac {\gamma}{\alpha})^\frac 13 = 1$
If $(\frac {\alpha}{\beta} + \frac {\beta}{\gamma} + \frac {\gamma}{\alpha}) = 3$
We can conclude:
$\frac {\alpha}{\beta} = \frac {\beta}{\gamma} = \frac {\gamma}{\alpha} = 1$
$2\log a = 3\log b = 5\log c\\ a^2 = b^3 = c^5$
$b = 9 = 3^2\\ a^2= b^3 = 3^6\\ a = 3^3\\ c = 3^{\frac 65}$