What is the solution set of $ \displaystyle\log _x\left({x+5\over 1-3x}\right)>0$?

Question

What is the solution set of $ \displaystyle\log _x\left({x+5\over 1-3x}\right)>0$?

I'm getting $0 < x < 1/3$ but that is wrong answer.

Answer 1

Let $\displaystyle y={x+5\over 1-3x}$. Then $\log_x y >0$ if $x>1$ and $y>1$, or if $0<x<1$ and $0<y<1$.

If $x>1$, then $y$ is a positive number divided by a negative number, so is negative.

Hence, we only need to consider when $x<1$. That means we must have $$0 < {x+5\over 1-3x} < 1.$$ If $x<{1\over3}$, then $1-3x>0$ and we get $$0<x+5 < 1-3x$$ which means $x>-5$ (automatic) and $4x < -4$ or $x<-1$ (impossible).

If $x>{1\over3}$, then $1-3x<0$, and we get $$0 > x+5 > 1-3x$$ which means $x<-5$ (impossible).

Thus there are NO values of $x$ where $\displaystyle\log _x {x+5\over 1-3x}>0$!

Answer 2

See we can re arrange our inequation as $$\frac{x+5}{1-3x}<1$$ so it becomes $x+5<1-3x$ reversal of equality takes place due to log. But there exists no positive x such that $x+5<1-3x$ note we can take $0$ as log to base $0$ isnt defined. Thus there is no x satisfying the equation