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how to solve e^-x = (e^x)^3 * 1/e^2
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ln of both sides -x = ln(e^(3x)/e^2) = ln(e^(3x) - ln(e^2) -x = 3x - 2 3x+x = 2 4x=2 x = (1/2)
\[e^{-x} = e{^x{^\left(3\right)}} \dot\ \frac{1}{e^2}\] \[\frac{1}{e^x}= e{^x{^\left(3\right)}} \dot\ \frac{1}{e^2}\] \[\frac{1}{e^x}= \frac{e{^x{^\left(3\right)}}}{e^2}\] \[e^2= e{^x{^\left(3\right)}}*e^x\] \[e^2= e{^x{^\left(3 + 1\right)}}\] \[e^2= e{^x{^\left(4\right)}}\] \[e^2= e{^{4x}}\] \[2 = 4x\] \[\frac{2}{4} = x\] \[\frac{1}{2} = x\]
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