<s>[INST] logarithm inequality: 2log5(x) - logx(125) I was given the following problem:
$2\log_{5}x$ - $\log_{x}125$ $< 1$
I tried to use the exponent law and the change of base law and I got this.
$\frac{\log_{}x^2}{\log_{}5}$ - $\frac{3log_{}5}{\log_{}x}$ $< 1$
Which I then changed back to:
$\frac{2log_{}x}{\log_{}5}$ - $\frac{3log_{}5}{\log_{}x}$ $< 1$
I then made 
$\frac{(log_{}x^2)(log_{}x)}{(log_{}5)(log_{}x)}$ - $\frac{(log_{}5)(log_{}125)}{(log_{}5)(log_{}x)}$ $< 1$
When I combined these, I got $\frac{(log_{}\frac{x^3}{625})}{log_{}5x}$ $< 1$, which I thought I could combine to: 
${log_{}\frac{x^2}{3125}}$ $< 1$
I got from here $x^2$ < 3,1250, which means x < $\sqrt31250$. However, the answer given is:
0 < x < 0.2 and 1 < x < 5$\sqrt5$.
Where did I go wrong? Does it have to do with original different basis, one being a constant and one being a variable? Can someone please help me solve this problem?
 [/INST] $2\log_{5}x - \log_{x}125 < 1$
$2\log_{5}x - 3\log_{x}5 < 1$
$2\log_{5}x - 3\frac{1}{\log_{5}x} < 1$At this point you need to be careful as $\log_5 x$ can be negative (which is where your mistake is).
If $\log_5 x > 0$ we have $2(\log_{5}x)^2 - \log_5 x -3 <0 \rightarrow 0 < \log_5 x < 1.5 \rightarrow 1 < x < 5\sqrt{5}$
If $\log_5 x < 0$ we have $2(\log_{5}x)^2 - \log_5 x -3 >0 \rightarrow  \log_5 x < -1 \rightarrow 0 < x < 1/5$
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