How do I express $(2^x)-1=2/(2^x)$ as a quadratic equation, and show that only one real solution, $x=1$ , exists?

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Answer 1 · 2018-02-21T20:20:58

See explanation...

Given:

$2^x-1 = 2/(2^x)$

Multiply both sides by $2^x$ to get:

$(2^x)^2-(2^x) = 2$

This is a quadratic equation in $2^x$ . If you wish you can use a substitution $t = 2^x$ to get $t^2-t=2$ , but I will just leave it as $2^x$ ...

Subtract $2$ from both sides to get:

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

So:

$2^x = 2" "$ or $" "2^x = -1$

Note that for any real value of $x$ we have $2^x > 0$

Hence the only possible real solutions are given by:

$2^x = 2$

The function $f(x) = 2^x$ is strictly monotonically increasing and therefore one to one.

We find:

$2^(color(red)(1)) = 2$

So $x=1$ is the unique real solution.

How do I express (2^x)-1=2/(2^x)(2^x)-1=2/(2^x) as a quadratic equation, and show that only one real solution, x=1x=1, exists?