Logarithm problem with two bases

Question

Given $$\log_x9+\log_9x=\dfrac{10}3.$$

How can I find the greatest value of $x$ that satisfies the equation above?

Answer 1

HINT:

$$\log_x9+\log_9x=\dfrac{10}3$$

Now by Change of Base of Logarithm use $\log_x9=\dfrac{\log9}{\log x}=\dfrac1{\dfrac{\log x}{\log9}}=\dfrac1{\log_9x}$ to form a Quadratic equation in $\log_9x$

Now if $\log_9x=y,x=9^y>0$ for real finite $y$

Answer 2

hint: $a = \log_{9}x$, then you have a quadratic in $a$: $a + \dfrac{1}{a} = \dfrac{10}{3}$.

Answer 3

Let $\log_x9=t;\frac1{\log_x9}=\frac1t$
Solve $t+\frac1t=\frac{10}3$
$3t^2+3=10t$
$3t^2-10t+3=0$
$t=\frac{10\pm\sqrt{100-4.3.3}}{6}=3$ or $\frac13$.
So $\log_x9=3$ or $\frac13$
Solve them, you will not get a neat answer, but an answer.