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

How do you solve for x in 2e^(x-2)=e^x + 72e^(x-2)=e^x + 7?