How can I find $x$ from this logarithms equation?

Question

Solve $\log_9x + \log_{81}3x = 1$

Answer 1

$\log_9x=2\log_{81}x$

So, $$81^{2\log_{81}x+\log_{81}3x}=81$$$$x^2\cdot3x=81$$$$x^3=27$$

So, if $x\in\mathbb R, x=3$

Answer 2

Hint: $\log_9 x= \log_{81}x^2$

Answer 3

Well, $\log_{81} a$ is the number you have to raise $81$ to, in order to get $a$. So to get $a$, you'd have to raise $9$ to twice that. So $\log_9 a = 2\log_{81} a.$ There's a general rule here that you could probably write down, if you think about it.

Using this idea, we can rewrite your equation:

$$\log_9 x + \frac{1}{2}\log_9 3x = 1.$$

Answer 4

Hint:

Make use of the fact that $a\log b=\log b^a$ and the fact that $\log_{a^n}b=1/n\cdot \log_ab$ and you'll be good to go.

$$\begin{align*}\log_9x+\log_{81}3x=1\\ \log_9x+\dfrac{1}{2}\log_93x=1\\ \log_9\sqrt{3}x^{3/2}=1\\ \sqrt{3}x^{3/2}=9\end{align*}$$

Can you proceed?