Solving $\log_{2x}2 = \log_{3x}3$, I find that $x$ can be any number. Did I do something wrong?

Question

Given the question

Find all possible values of $x$ when $\log_{2x}2 = \log_{3x}3$

This is how I did it..

$\log_{2x}2 = \log_{3x}3$

$1/ \log_2 2x = 1/ \log_3 3x$

Therefore $\log_2 2x = \log_3 3x$

$\log_2 2 + \log_2 x = \log_3 3 + \log_3 x$

$\log_2 x = \log_3 x$

$\frac{\log_2 x}{\log_3 x} = 1$

$\log_x x = 1$????

So x can be any number?

Can someone explain this a pre-calculus level? Thanks.

Answer 1

I do not know how do you obtain $\log_x(x)=1$.

From $\log_2 (x) = \log_3(x)$.

We have $$\frac{\ln x}{\ln 2} = \frac{\ln x}{\ln 3}$$

$$\ln x \left(\frac1{\ln 2} - \frac1{\ln 3} \right)=0$$

Hence $\ln x = 0$, hence $x=1$.

Answer 2

How exactly did you perform the following step? It’s clearly not correct:

$$\color{red}{\frac{\log_2 x}{\log_3 x} = 1 \iff \log_x x = 1}$$

Instead, you must simplify as follows:

$$\frac{\log_2 x}{\log_3 x} = 1 \iff \frac{\frac{\ln x}{\ln 2}}{\frac{\ln x}{\ln 3}} = 1 \iff \frac{\ln x}{\ln 2} = \frac{\ln x}{\ln 3} \iff x = 1$$