MHB 6.6.63 ln(7-x)+ln(1-x)=ln(25-x)

[karush]

$\tiny\textbf{6.6.63 Kiliani HS}$
Solve for x give exact form
$\ln{(7-x)}+\ln{(1-x)}=\ln{(25-x)}$

$\begin{array}{rrll}
\textsf{log rules} &(7-x)(1-x) &=25-x \\
\textsf{expand} &7-8x+x^2 &=25-x \\
\textsf{set to zero} &x^2-7x-18 &=0 \\
\textsf{factor} &(x-9)(x+2) &=0 \\
\textsf{zero's} &x&=9, \quad -2 \\
x\le -1\quad\therefore &x&=-2
\end{array}$

I hope...
typo's ?
have to very careful

 

[DaalChawal]

there will be $x \lt 1$ not -1 so $x=-2$ will be the answer