Solve for x to three significant digits?

Question

$e^x + e^-x = 2$

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Answer 1 · 2016-12-13T15:14:45

Real solution: $x = 0$

Complex solutions: $x = 2kpi i$ for any integer $k$

Let $t = e^x$

Then our equation becomes:

$t + 1/t = 2$

Mutliply through by $t$ and subtract $2t$ from both sides to find:

$0 = t^2-2t+1 = (t-1)^2$

So the only possible value of $t$ is $t = 1$

Now solve $e^x = 1$

The unique Real solution is $x = 0$ since $a^0 = 1$ for all $a != 0$

There are also Complex solutions resulting from Euler's identity:

$e^(ipi) = -1$

Hence:

$e^((2kpi)i) = ((e^(ipi))^2)^k = ((-1)^2)^k = 1^k = 1$

for any integer $k$