How do you solve $2 e^{x} - 1 = 0$ $2e^x-1=0$ ?

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Answer 1 · 2018-06-11T20:43:51

$x = ln (\frac{1}{2})$ $x= ln(1/2)$

$2 e^{x} - 1 = 0$ $2e^x-1=0$

$2 e^{x} = 1$ $2e^x=1$

$e^{x} = \frac{1}{2}$ $e^x=1/2$

Take natural log of both sides:
${ln e}^{x} = ln (\frac{1}{2})$ $lne^x=ln(1/2)$

$x = ln (\frac{1}{2})$ $x= ln(1/2)$

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