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0.5e^0.08x=40
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\[\large \frac{1}{2}e^{0.08x}=40\]Start by multiplying both sides by 2,\[\large \cancel{2}\cancel{\frac{1}{2}}e^{0.08x}=40(2)\] Giving us,\[\large e^{0.08x}=80\] Then we'll take the natural log of both sides,\[\large \ln\left(e^{\color{royalblue}{0.08x}}\right)=\ln(80)\]
The log allows us to get the x out of the exponent position. Here is an important log rule,\[\large \log(a^{\color{orangered}{b}})=\color{orangered}{b}\log(a)\]
Applying this to our problem gives us,\[\large \color{royalblue}{0.08x}\ln\left(e\right)=\ln(80)\]
Then just do a few algebra steps to solve for x!
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