natural log problem
- Thread starter lowend
- Start date
Hello, Steve!
\(\displaystyle \text{We have: }\;12 + 20\ln x \:=\:90 \quad\Rightarrow\quad 20\ln x \:=\:78 \quad\Rightarrow\quad \ln x \,=\,\tfrac{78}{20} \,=\,3.9\)
\(\displaystyle \text{Therefore: }\;x \:=\:e^{3.9} \;\approx\;49.4\)
\(\displaystyle \text{Given: }\\,=\, 12 + 20\ln x,\,\text{ and }P = 90\)
\(\displaystyle \text{Solve for }x.\)
\(\displaystyle \text{We have: }\;12 + 20\ln x \:=\:90 \quad\Rightarrow\quad 20\ln x \:=\:78 \quad\Rightarrow\quad \ln x \,=\,\tfrac{78}{20} \,=\,3.9\)
\(\displaystyle \text{Therefore: }\;x \:=\:e^{3.9} \;\approx\;49.4\)