Solving $\ln\left(\frac{\ x-2}{x-3}\right)=2$

Question

Question:

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

Workings:

$\log_{x-3}(x-2)=2$
$(x-2)=(x-3)^2$
$(x-2)=(x^2-6x+9)$
$x^2-7x+11=0$

Use -b formula

and arrive at $x=\frac{7+\sqrt 5}{2}$ and $x=\frac{7-\sqrt 5}{2}$

how's my workings holding up here?

Answer 1

Remember that $\ln{e}=1$ and $a\ln{x}=\ln{x^a}$. Therefore:

$$2=2\cdot 1=2\cdot\ln{e}=\ln{e^2}.$$

So, what we have now is:

$$ \ln{\frac{x-2}{x-3}}=\ln{e^2}. $$

It's easy to see that the two sides of the equation are gong to be equal to each other only if the expressions inside the logarithms are equal. Thus, the previous equation is equivalent to this new one with the following restriction $\frac{x-2}{x-3}>0\implies x\in(-\infty,2)\cup(3,\infty)$:

$$ \frac{x-2}{x-3}=e^2. $$

The solution to this equation with that domain restriction will be the solution to your original equation.

Answer 2

It is not holding well. In fact, $\ln\left(\frac{x-2}{x-3}\right)=2\iff\frac{x-2}{x-3}=e^2$. And $\ln\left(\frac{x-2}{x-3}\right)\neq\ln_{x-3}(x-2)$.