## Calculus 8th Edition

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