Find the range of values that $x$ can take if $9 \log_x5 = \log_5x$ I'm stuck on a homework question about logarithms. I can't work out how to do it, and all I've managed to do is turn $9 \log_x5$ into $ \log_x5^9$. Can anyone guide me onto the right path to solve this?
 A: Using $\log_ab=\dfrac{\log b}{\log a},$
we have $$(\log x)^2=9(\log 5)^2=(3\log5)^2=[\log(5^3)]^2$$
$$\implies\log x=\pm \log(125)$$
Either $$\log x=\log(125)$$
Or $$\log x=-\log(125)=(\log125^{-1})$$
Now for real positive $c,d;\log c=\log d\iff c=d$
A: $$\log_x 5= \frac{\log_5 5}{\log_5 x} \Rightarrow \log_x 5=\frac{1}{\log_5 x}$$
$$9 \log_x 5= \log_5 x \Rightarrow  \frac{9}{\log_5 x}=\log_5 x \\ \Rightarrow (\log_5 x)^2=9 \Rightarrow \log_5 x= \pm 3 \Rightarrow 5^{\log_5 x}=5^{\pm 3} \Rightarrow x=5^{\pm 3}$$
So:
$$x=125 \text{ or } x=\frac{1}{125}$$
A: If you prefer to use powers rather than logs you can proceed in this way:
Let $L=\log_5 x$ so that $x=5^L$ and $5=x^{\frac 1L}$ so that $\log_x5=\frac 1L$
We then have $9\cdot \frac 1L = L$ so that $9=L^2$ and $L=\pm 3$ so that $x=5^3$ or $x=5^{-3}$
A: We have
$$9\frac{\ln(5)}{\ln(x)}=\frac{\ln(x)}{\ln(5)}$$ if $x>0$ and $x\ne 1$.       
A: For me, it is more natural to go through $\cdots$ natural logarithms. So, $$9log_x5 - log_5x=0$$ becomes $$\frac{9 \log (5)}{\log (x)}-\frac{\log (x)}{\log (5)}=0$$ which simplifies to $$9 \log ^2(5)-\log ^2(x)=0$$ that is to say $$\log(x)=\pm 3\log(5)=\pm \log(5^3)=\pm \log(125)$$ and then the answers you already received.
For the more general problem  $$k\log_xa - \log_ax=0$$ the same procedure would lead to $$\log(x)=\pm \sqrt k \log(a)=\pm \log(a^{ \sqrt k })$$ that is to say $$x=a^{\pm \sqrt k }$$
