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

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Answer 1 · 2015-10-14T23:04:32

$x = ln(5)$

Multiply both sides by $e^x$

$e^(x+x) - 15e^(x-x) - 2e^x = 0$
$e^(2x) - 15 - 2e^x = 0$

Call $e^x = y$ so we have

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

Solve the quadratic

$y = (2 +-sqrt(4 - 4*1*(-15)))/2$

$y = (2 +-sqrt(4 +60))/2$

$y = (2 +- 8)/2$

$y = 1 +- 4$

$y = 5$ 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

$x = ln(5)$

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