$$ \log_{x}{(x+3)} > \log_{x}{(2x)} $$ My work so far is
Step 1 : Finding the definition term for base of logarithm $$ x > 0 \lor x \neq 1,x \in Real $$ Step 2 : Finding the definition term for $\log_{x}{(x+3)}$ $$ x+3>0 $$ $$x>-3 $$ Step 3 : Finding the definition term for $\log_{x}{(2x)}$ $$ 2x>0$$ $$ x>0 $$ Step 4 : Unify the definition term from step 1 to step 3 $$ x > -3 \lor x >0 \lor x\neq 1 $$ $$\implies x>0 \lor x\neq 1 $$ Step 5 : Finding the inequality term I change the base of logarithm, for apllying the cancelation of "logarithm with x as its base" $$ \frac{\log_{10}{(x+3)}}{\log_{10}{(x)}} > \frac{\log_{10}{(2x)}}{\log_{10}{(x)}}$$ I try multiply both side with $\log_{10}{(x)}$ $$ \log_{10}{(x+3)} > \log_{10}{(2x)} $$ $$ \iff x+3 > 2x$$ $$x-2x>-3$$ $$-x>-3$$ Multiply with –1 $$x<3$$ Step 6 : Finding the solution set $$ = \{ x | 0<x<3, x \neq 1, x \in Real \}$$
I just finish it or i missed something to solve this inequality ? Please, give your suggestion and help to explain. Thank you so much.
It may be easier to proceed as follows.
Let $a= \log_{x}{(x+3)}$ and $b=\log_{x}{(2x)}$. Then,
$$a>b,\>\>\>\>\> x^a=x+3,\>\>\>\>\>x^b = 2x$$
Two cases:
1) $x>1 \implies x^a>x^b \implies x+3>2x$, which leads to the solution $1<x<3$.
2) $0<x<1 \implies x^a<x^b \implies x+3<2x$, which has no solution.
Thus, $1<x<3$.