How do you solve $e^x + e^(-x) = 3$ ?

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Answer 1 · 2016-01-01T18:02:37

Express as a quadratic in $t = e^x$ , solve and take logs to find:

$x = ln((3+-sqrt(5))/2) =+-ln((3+sqrt(5))/2)$

Let $t = e^x$ .

Then the equation becomes:

$t + 1/t = 3$

Multiplying both sides by $t$ we get:

$t^2+1 = 3t$

Subtract $3t$ from both sides to get:

$t^2-3t+1 = 0$

$t = (3+-sqrt(5))/2$

Note that due to the symmetry of the equation $t+1/t = 3$ in $t$ and $1/t$ , these two values are actually reciprocals of one another.

Now $t = e^x$ , so:

$e^x = (3+-sqrt(5))/2$

Taking natural logs of both sides we find:

$x = ln((3+-sqrt(5))/2) =+-ln((3+sqrt(5))/2)$

How do you solve e^x + e^(-x) = 3e^x + e^(-x) = 3?