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?

you can get rid of the natural logarithm by applying $e^{ln(\frac{x-2}{x-3})} = e^2$ — Donlans Donlans, Aug 16 '19 at 22:29
Actually, $;\log_{x-3}(x-2)$ (which is defined only if $x>3$) is equal to $;\dfrac{\ln(x-2)}{\ln(x-3)}$. — Bernard, Aug 16 '19 at 22:56

score 2 · Answer 1 · answered Aug 16 '19 at 22:27

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 $\ln\left(\frac{x-2}{x-3}\right)\neq\ln_{x-3}(x-2)$.

answered Aug 16 '19 at 22:27

José Carlos Santos

427,504

1

Okay, thank you. I'll work on that now – Alan Marsh Aug 16 '19 at 22:28

Michael Rybkin · Accepted Answer · 2019-08-17T00:47:41.010

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.

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

2 Answers2