## Algebra 2 (1st Edition)

$\ln(10x)$
We know that $x\log_ab=\log_ab^x$ and that $\log_ab+\log_ac=\log_a(bc)$ Hence here $\ln40+2\ln0.5+\ln x=\ln40+\ln0.5^2+\ln x=\ln40+\ln0.25+\ln x=\ln(40\cdot0.25x)=\ln(10x)$