Answer
$x=5$ or $x=-2$
Work Step by Step
Recall the definition of a logarithm (pg. 451):
$\log_{b}{x}=y$ iff $b^y=x$
Applying this definition, we get:
$7^2=(2x-3)^2$
$49=(2x-3)^2$
$\pm \sqrt{49}=2x-3$
$\pm 7=2x-3$
$\pm 7+3=2x$
$\dfrac{\pm 7+3}{2}=x$
$x=\frac{7+3}{2}=5$ or $x=\frac{-7+3}{2}=-2$
We confirm that the answers work:
$\log_7 (2\cdot5-3)^2=2$
$\log_7 (10-3)^2=2$
$\log_7 (7)^2=2$
$\log_7 (7)^2=2$
$2=2$
$\log_7 [2\cdot (-2)-3]^2=2$
$\log_7 (-4-3)^2=2$
$\log_7 (-7)^2=2$
$\log_7 7^2=2$
$2=2$
