Logarithm problem with two bases Given $$\log_x9+\log_9x=\dfrac{10}3.$$

How can I find the greatest value of $x$ that satisfies the equation above?
hint: $a = \log_{9}x$, then you have a quadratic in $a$: $a + \dfrac{1}{a} = \dfrac{10}{3}$.
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 $\log_9x=y,x=9^y&gt;0$ for real finite $y$