Solving an inequality with logarithms of different bases?

Question

The question is to solve $\log_2x>\log_3x$ for x.

I attempted to simplify it like so: \begin{align*} \log_2x&>\log_3x\\ \frac{\log_2x}{\log_22}&>\frac{\log_2x}{\log_23}\\ \log_2x&>\log_2x^{\frac{1}{\log_23}}\\ x&>x^{\frac{1}{\log_23}} \end{align*}

It feels like I'm really close but not sure how to finish this problem.

Answer 1

Try the steps below:

$$\log_2x >\log_3x$$ $$\frac{\ln x}{\ln2}>\frac{\ln x}{\ln 3}$$ $$\ln x \left(\frac{1}{\ln2} - \frac{1}{\ln3} \right)>0$$

Since

$$\frac{1}{\ln2} - \frac{1}{\ln3} >0$$

the following must hold,

$$\ln x > 0$$

Thus,

$$x>1$$

Answer 2

This solution uses the same technique you began to use.

Try dividing both sides by $x$. We can do this since taking the logarithm assumes anyway $x\not=0$. You will get:

$$1>x^{\log_3(2) -1}$$

Now find out if $\log_3(2) -1$ is positive or negative and you should be able to finish from there!