Answer
(a) $x=(\frac{7-ln6}{4})$
(b) $x=\frac{1}{3}(10+e^{2})$
Work Step by Step
(a) Given: $e^{7-4x}=6$
Take logarithms to both sides.
$(7-4x)=ln6$
This implies
$x=(\frac{7-ln6}{4})$
(b) Given: $ln(3x-10)=2$
$3x-10=e^{2}$
$x=\frac{1}{3}(10+e^{2})$