How do you solve for x in #2e^(x-2)=e^x + 7#?

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Answer 1 · 2016-06-14T05:58:01

#x=ln7-ln(2e^(-2)-1)#

#2e^(x-2)=e^x+7#

or #2e^x/e^2=e^x+7#

or #2e^x/e^2-e^x=7#

or #e^x(2e^(-2)-1)=7#

or #e^x=7/((2e^(-2)-1)#

or #x=ln(7/((2e^(-2)-1)))#

or #x=ln7-ln(2e^(-2)-1)#