Answer
$x\approx -1.15$
Work Step by Step
Take the natural log of both sides to obtain:
$\ln{(4^{1-x})} = \ln{(3^{2x+5})}$
Use the rule $\log{(a^n)} = n\cdot\log{a}$ to obtain:
$(1-x)\ln{4} = (2x+5)\ln{3}$
Distribute $ln{4}$ on the left and $\ln{3}$ on the right to obtain:
$\ln{4} - x\ln{4} = 2x\ln{3} + 5\ln{3}$
Subtract $5\ln3$ on both sides of the equation to obtain:
$\ln4 - x\ln4 - 5\ln3 = 2x\ln3+5\ln3 - 5\ln3
\\\ln4 - x\ln4 - 5\ln3 = 2x\ln3$
Add $x\ln4$ to both sides of the equation to obtain:
$\ln4 - x\ln4 - 5\ln3 +x\ln4= 2x\ln3+x\ln4
\\\ln4 - 5\ln3 = 2x\ln3+x\ln4$
Factor out $x$ on the right side of the equation to obtain:
$\ln4 -5\ln3 = x(2\ln3+\ln4)$
Divide $\ln3+\ln4$ to both sides of the equation to obtain:
$\dfrac{\ln4-5\ln3}{2\ln3 + \ln4}=\dfrac{x(2\ln3+\ln4)}{2\ln3+\ln4}
\\\dfrac{\ln4-5\ln3}{2\ln3 + \ln4}=x$
Use a calculator to obtain:
$x\approx -1.15$
