How do you solve e^x - 15e^(-x) - 2 = 0 ?

Oct 14, 2015

$x = \ln \left(5\right)$

Explanation:

Multiply both sides by ${e}^{x}$

${e}^{x + x} - 15 {e}^{x - x} - 2 {e}^{x} = 0$
${e}^{2 x} - 15 - 2 {e}^{x} = 0$

Call ${e}^{x} = y$ so we have

${y}^{2} - 15 - 2 y = 0$
${y}^{2} - 2 y - 15 = 0$

$y = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot \left(- 15\right)}}{2}$

$y = \frac{2 \pm \sqrt{4 + 60}}{2}$

$y = \frac{2 \pm 8}{2}$

$y = 1 \pm 4$

$y = 5$ or $y = - 3$

But $y = {e}^{x}$ so we really have

${e}^{x} = 5$ or ${e}^{x} = - 3$

The exponential function is positive for every real $x$ so we can ignore that root and say that

${e}^{x} = 5$

Taking the log of both sides we have

$x = \ln \left(5\right)$