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
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.