# Thread: Logarithmic equation help

1. ## Logarithmic equation help

log(x-1) - log(x+6) = log(x-2) - log(x-3)

Is this right?

(x-1)/(x+6) = (x-2)/(x-3)

(x-3)(x-1)/(x+3)(x+6) = (x-2)(x+6)/(x+3)(x+6)

(x-1)/(x-6) = (x-2)/(x-3)

x^2 + 4x - 12 = x^2 - 4x - 3

x = 9/8

2. Looks right. BUT
log(x-2)=log(9/8-2)=log(-7/8) which is not defined.

So, something's wrong !

-O

3. Originally Posted by NeedHelp18
log(x-1) - log(x+6) = log(x-2) - log(x-3)

Is this right?

(x-1)/(x+6) = (x-2)/(x-3)

(x-3)(x-1)/(x+3)(x+6) = (x-2)(x+6)/(x+3)(x+6)

(x-1)/(x-6) = (x-2)/(x-3)
huh?

shouldn't that line be (x - 1)(x - 3) = (x - 2)(x + 6)? ...(which would yield (x - 1)/(x + 6) = (x - 2)/(x - 3) ...but that's not what we would do)

expand both sides:

$\displaystyle x^2 - 4x ~{\color{red}+}~ 3 = x^2 + 4x - 12$

now solve

4. Hello, NeedHelp18!

$\displaystyle \log(x-1) - \log(x+6) \:=\: \log(x-2) - \log(x-3)$

Is this right?

$\displaystyle \log\left[\frac{x-1}{x+6}\right] \:=\:\log\left[\frac{x-2}{x-3}\right]$

$\displaystyle \frac{x-1}{x+6} \:=\:\frac{x-2}{x-3}$

$\displaystyle \frac{{\color{red}\rlap{////////////}}(x-3)(x-1)}{{\color{red}\rlap{////////////}}(x+3)(x+6)} \:{\color{red}\rlap{///}}=\: \frac{{\color{red}\rlap{////////////}}(x-2)(x+6)}{{\color{red}\rlap{////////////}}(x+3)(x+6)}$ .
. . . What was all that about?

$\displaystyle \frac{x-1}{x+6} \:=\:\frac{x-2}{x-3}$

$\displaystyle x^2 + 4x - 12 \:= \:x^2 - 4x {\color{red}\rlap{\;///}}- 3$ . . . . no

$\displaystyle x \:= \:\frac{9}{8}$

We have: .$\displaystyle \log(x-1) - \log(x+6) \:=\:\log(x-2)-\log(x-3)$

Rearrange terms: .$\displaystyle \log(x-1) + \log(x-3) \:=\:\log(x-2) + \log(x+6)$

And we have: .$\displaystyle \log\bigg[(x-1)(x-3)\bigg] \:=\:\log\bigg[(x-2)(x+6)\bigg]$

Exponentiate: .$\displaystyle (x-1)(x-3) \:=\:(x-2)(x+6) \quad\Rightarrow\quad x^2-4x+3 \:=\:x^2+4x-12$

. . which simplifies to: .$\displaystyle 8x \:=\:15 \quad\Rightarrow\quad x \:=\:\frac{15}{8}$

But this answer turns out to be extraneous.

The equation has no solution.