Answer
no solution
Work Step by Step
RECALL:
(1) $\ln{P} + \ln{Q}=\ln{(PQ)}$
(2) $\ln{P}=\ln{Q} \longrightarrow P=Q$
Use rule (1) above to obtain
$\ln{[(x-2)(3)]}=\ln{(5x-7)}$
Use rule (2) above to obtain:
$(x-2)(3)=5x-7
\\x(3) -2(3) = 5x-7
\\3x-6=5x-7$
Add $7$ and subtract$3x$ on both sides of the equation to obtain:
$3x-6+7-3x=5x-7+7-3x
\\(3x-3x) -6+7=(5x-3x) -7+7
\\1=2x$
Divide $2$ on both sides of the equation to obtain:
$\frac{1}{2}=x$
However, ,if $x=\frac{1}{2}$,
$\ln{(x-2)} = \ln{(\frac{1}{2}-2)} = \ln{(-\frac{3}{2})}$ which is undefined.
$\ln{(5-7)} = \ln{(5\cdot\frac{1}{2} - 7)} = \ln{(\frac{5}{2} - \frac{14}{2})} = \ln{(-\frac{9}{2})}$ which is undefined.
Thus, the given equation has no solution.
