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e^(x+2) = e^(x) + 5
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\[e^{x+2}=e^2e^x\] should be a good start
\[e^2e^x-e^x=5\] \[e^x(e^2-1)=5\] \[e^x=\frac{5}{e^2-1}\]
take the log to get \[x=\ln(\frac{5}{e^2-1})\]
k?
great! thank satellite783
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