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e^(x+6)=(e^x)+6
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Ooo that 6 sure is sneaky, huh?
hmm
\[\Large\rm e^{x+6}=e^x\cdot e^6\]There our problem:\[\Large\rm e^{x+6}=e^x+6\]Can be written as:\[\Large\rm e^x\cdot e^6=e^x+6\]\[\Large\rm e^x\cdot e^6-e^x=6\]\[\Large\rm e^x(e^6-1)=6\]
\[\Large\rm e^x=\frac{6}{e^6-1}\]Then we can use a log to get the x out of the exponent position, yah?
`There our problem`? That was a weird typo lol
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not sure how to do that @zepdrix
i also got up to the part that you did but got stuck after
\(\large\tt \begin{align} \color{black}{ e^x=\frac{6}{e^6-1}\\~\\ \ln e^x=\ln (\frac{6}{e^6-1})\\~\\ x\ln e=\ln (\frac{6}{e^6-1})\\~\\ x=\ln (\frac{6}{e^6-1})-----(use~~e=2.718)\\~\\ x=\ln(0.0149.....)\\~\\ x=-4.2}\end{align}\)
thank you so much
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