logarithm inequality: 2log5(x) - logx(125)

Question

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?

Answer 1

If you call $y=log_5 x$, then your expression turns into $2y-3/y<1$. Try continue from here

Answer 2

$2\log_{5}x - \log_{x}125 < 1$
$2\log_{5}x - 3\log_{x}5 < 1$
$2\log_{5}x - 3\frac{1}{\log_{5}x} < 1$
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 $2(\log_{5}x)^2 - \log_5 x -3 >0 \rightarrow \log_5 x < -1 \rightarrow 0 < x < 1/5$

Answer 3

To address the explicit question, "Where did I go wrong?," you seem to be assuming that $(\log x^2)(\log x)=\log x^3$, $(\log5)(\log125)=\log625$, and $(\log5)(\log x)=\log5x$, which would all be correct if $(\log a)(\log b)$ were $\log ab$. But the correct logarithmic identity is $\log ab=\log a+\log b$, not $\log a\times\log b$.

Answer 4

Up to this line it is correct: $$\frac{(\ln x^2)(\ln x)}{(\ln 5)(\ln x)}-\frac{(\ln 5)(\ln 125)}{(\ln 5)(\ln x)}<1$$ Starting from next line you made two mistakes: $$1) \ \ (\ln x^2)\cdot (\ln x) \ne \ln x^3 \quad (\ln x^2+\ln x=\ln x^3)\\ 2) \ \ \frac{\ln \frac{x^3}{625}}{\ln (5x)}\ne {\ln \frac{x^2}{3125}} \quad (\ln \frac{x^3}{625}-\ln (5x)=\frac{x^2}{3125})$$ Instead, continue as follows: $$\frac{2(\ln x)(\ln x)}{(\ln 5)(\ln x)}-\frac{3(\ln 5)(\ln 5)}{(\ln 5)(\ln x)}-\frac{(\ln 5)(\ln x)}{(\ln 5)(\ln x)}<0 \Rightarrow \\ \frac{2(\ln x)^2-(\ln 5)(\ln x)-3(\ln 5)^2}{(\ln 5)(\ln x)}<0 \Rightarrow \\ \frac{(\ln x+\ln 5)(2\ln x-3\ln 5)}{(\ln 5)(\ln x)}<0 \Rightarrow \\ \text{the zeros are:} \quad x=\frac15,1,5^{3/2}, \text{also $x>0$, so:} \\ x\in(0,\frac15)\cup (1,5^{3/2}).$$