# Solving the exponential equation $2e^x (x-1)=0$

How does one solve the equation $2e^x (x-1)=0$ ?

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Is that $2(x-1)e^x = 0$ or $2e^{x(x-1)} = 0$? Or something else? –  Aryabhata Jan 26 '11 at 2:33

If $y$ and $z$ are Real (Complex) numbers, then $yz = 0$ exactly when at least one of $y = 0$ or $z = 0$. Since $x \mapsto 2e^x$ and $x \mapsto x-1$ both are maps from Real numbers to Real numbers, $2e^x(x-1) = 0$ will be true exactly when at least one of the following is true:

(1) $2e^x = 0$ $\Leftrightarrow$ $2 = 0$ or $e^x = 0$ (both impossible)

(2) $x - 1 = 0$ $\Leftrightarrow$ $x = 1$

so your only solution is $x = 1$.

Note that this is the same reason we factor the quadratic polynomial in the equation $ax^2 + bx +c = 0$, mainly so that we can break up the equation into two linear ones that are easily solved.

Also since I've seen this done before, $yz = r$ does NOT necessarily mean that either $y = r$ or $z = r$ when $r \neq 0$. The reason $0$ holds a special place here is because if say $y \neq 0$, then we may multiply both sides by $1/y$ to get $z = r/y$, which is of course $0$ exactly when $r = 0$.

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Note that $e^x$ is always positive. So if your equation is $e^{x(x-1)} = 0$ then it has no solution. (Not over the reals, that is.)

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Even over complex numbers, $e^z$ is never zero. –  Aryabhata Jan 26 '11 at 2:56
Good point actually. I love that function. –  Myself Jan 26 '11 at 12:18