How do you solve $e^{x} - 9 x e^{x} = 0$ $e^x-9xe^x=0$ ?

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Answer 1 · 2015-09-01T15:15:53

Factor and solve each factor equal to $0$ $0$

$e^{x} - 9 x e^{x} = 0$ $e^x-9xe^x=0$

$e^{x} (1 - 9 x) = 0$ $e^x(1-9x) = 0$

$e^{x} = 0$ $e^x = 0$ $" "$ or $" "$ $1 - 9 x = 0$ $1-9x =0$

$e^{x}$ $e^x$ is never $0$ $0$

$1 - 9 x = 0$ $1-9x =0$ when $= = \frac{1}{9}$ $= =1/9$

The only solution is $\frac{1}{9}$ $1/9$

How do you solve e^x-9xe^x=0ex−9xex=0e^x-9xe^x=0?