Solve for: ln(ax+b)-ln(cx+1)+e^3=5

Expression: $\ln(ax+b)-\ln(cx+1)+e^{3}=5$

Add $\ln(cx+1)$ to both sides

$\ln(ax+b)-\ln(cx+1)+e^{3}+\ln(cx+1)=5+\ln(cx+1)$

Simplify

$e^{3}+\ln(ax+b)=5+\ln(cx+1)$

Apply log rules

$e^{e^{3}}(ax+b)=e^{5}(cx+1)$

Solve $ e^{e^{3}}(ax+b)=e^{5}(cx+1):{\quad}x=\frac{e^{5}-e^{e^{3}}b}{e^{e^{3}}a-e^{5}c}$

$x=\frac{e^{5}-e^{e^{3}}b}{e^{e^{3}}a-e^{5}c}$