Find range of $x$ if $\log_5\left(6+\frac{2}{x}\right)+\log_{1/5}\left(1+\frac{x}{10}\right)\leq1$

Question

If $\log_5\left(6+\dfrac{2}{x}\right)+\log_{1/5}\left(1+\dfrac{x}{10}\right)\leq1$, then $x$ lies in _______

My Attempt

$$ \log_5\bigg(6+\dfrac{2}{x}\bigg)+\log_{1/5}\bigg(1+\dfrac{x}{10}\bigg)=\log_5\bigg(6+\dfrac{2}{x}\bigg)-\log_{5}\bigg(1+\dfrac{x}{10}\bigg)\leq1\\ \log_5\frac{(6x+2)10}{x(10+x)}\leq1\implies\frac{(6x+2)10}{x(10+x)}\leq5\\ \frac{4(3x+1)}{x^2+10x}\leq1\\ \implies 12x+4\leq x^2+10x\quad\text{or}\quad12x+4>x^2+10x\\ x^2-2x-4\geq0\quad\text{or}\quad x^2-2x-4<0\implies x\in\mathcal{R} $$

My reference gives the solution $(-\infty,1-\sqrt{5})\cup(1+\sqrt{5},\infty)$, what is going wrong here ?

Answer 1

You forgot to check when $$6+\dfrac{2}{x}>0$$ and $$1+\dfrac{x}{10}>0$$

is true!

Answer 2

By your work we need to solve $$\frac{x^2-2x-4}{x(x+10)}\geq0$$ and the domain gives $x>0$ or $-10<x<-\frac{1}{3}.$

The first by the interval's method gives $$1-\sqrt5\leq x<0$$ or $$x\geq1+\sqrt5$$ or $$x<-10,$$ which with our domain gives the answer: $$\left[1-\sqrt5,-\frac{1}{3}\right)\cup[1+\sqrt5,+\infty).$$

Answer 3

Writing your inequality in the form $$\frac{\ln\left(6+\frac{2}{x}\right)}{\ln(5)}+\frac{\ln\left(1+\frac{x}{10}\right)}{-\ln(5)}\le 1$$ we get the inequalities $$6+\frac{2}{x}>0$$ and $$1+\frac{x}{10}>0$$ and $$\frac{6+\frac{2}{x}}{1+\frac{x}{10}}\le 5$$ we get $$1-\sqrt{5}\le x<-\frac{1}{3}$$ or $$x\geq 1+\sqrt{5}$$