I am looking for a simple solution to this inequality:
$\log _x\left(3\right)<\log _{x+2}\left(9\right)$
I can solve it by switching to base 3, and move sides to get: $\frac{\log \:_3\left(\frac{x+2}{x^2}\right)}{\log _3\left(x\right)\cdot \log _3\left(x+2\right)} < 0$ and then take all cases where odd number of elements (out of the 3 logs) is negative.
But I wonder. Is there a solution without so many cases? i.e. just by basic log rules or with just two cases?
Thaks
On
We have $$\log_x3<\log_{x+2}3^2,$$ or $$\frac{2}{\log_3x}<\frac{1}{\log_3(x+2)}.$$
Now we only need consider the two cases $0<x<1$ and $x>1$ to proceed. For the last case, when both sides are positive, we obtain $$\log_3(x+2)^2<\log_3x,$$ from where you should be able to proceed. For the first case, when LHS is negative and RHS positive, we obtain $$\log_3x<\log_3(x+2)^2.$$ Again you should be able to proceed.
PS. You should see that the second case gives no solution, whereas the first is satisfied for all valid $x,$ that is $0<x<1.$
Use the chance of base rule which says $ \log_a b = \frac{ \log_{c} a } { \log_{c} b }$.
With that, your inequality becomes $ \frac{ \log 3 } { \log x } < \frac { \log 9 } { \log (x+2)} $
$ \Leftrightarrow \log^2 (x+2) \log x < 2 \log (x+2) \log^2 x $
$ \Leftrightarrow \log (x+2) \times \log x \times ( 2 \log x - \log(x+2)) > 0 $.
By checking the zeros, this has solution set $ 0 < x < 1 \cup 2 < x $.