How to solve $\log_{\frac{1}{3}}(\frac{3x-1}{x+2})>0$.

Question

How to solve

$$\log_{\frac{1}{3}}(\dfrac{3x-1}{x+2})>0$$

I am not sure how to solve it.

Answer 1

Hint: $$\log_{(1/3)}\left(\frac{3x-1}{x+2}\right)=-\log_{3}\left(\frac{3x-1}{x+2}\right)=\log_3\left(\frac{x+2}{3x-1}\right)>0.$$ Note that $\log_a t>0$ if and only if $t>1$ (when $a>1$).

Answer 2

Note: $$\log_{\frac13}(k)>0 \text{ when } 0<k<1$$

So you want when: $$0<\frac{3x-1}{x+2}<1$$ $$\to0^2<\frac{(3x-1)^2}{(x+2)^2}<1^2$$ $$\to0^2(x+2)^2<(3x-1)^2<1^2(x+2)^2$$ $$\to0<(3x-1)^2<(x+2)^2$$ Now continue this by finding all $x$ which satisfy: $$(x+2)^2>(3x-1)^2\text{ AND } (3x-1)^2>0$$