Find value of $x$ which satisfies the equation $\log_9 x=(\log_3 x)^2 ,x>1$

Question

So far I got this $(\log_3 x)/2=(\log_3 x)(\log_3 x)$. Then I am stuck. Any idea or nice elaboration on this problem will be kindly appreciated.

Answer 1

Well $$\log_ax=\frac{\log_b x}{\log_b a}.$$ Then $$\log_9x=\frac{\log_3 x}{\log_3 9}=\frac{\log_3 x}{2}.$$ Then by placing ${\log_3 x}=t$ you get the equation $$\frac{t}{2}=t^2.$$ From here either $t=0$ and $x=1$ or $t=\frac{1}{2}$ and then $x=\sqrt{3}$.

Answer 2

Note that: $$0=\log_9(x)-\left(\log_3(x)\right)^2 = \frac{\log(x)}{\log(9)}-\left(\frac{\log(x)}{\log(3)}\right)^2 = \log(x)\left(\frac{1}{\log(9)}-\frac{\log(x)}{\log(3)^2}\right). $$

Since $x>1$, $\log(x)>0$ and, hence, the second term needs to be 0: $$\frac{1}{\log(9)}-\frac{\log(x)}{\log(3)^2} =0. $$

Try to follow from there.

Answer 3

Well I would like to show one more way to do this :

We know that, $$\log_{a^b}c^d=\frac{d}{b}\log_{a}c$$ Using that here $\frac{1}{2}\log_3{x}=(\log_3{x})^2$. Then it is just a quadratic equation with $\log_3{x}=t$