What do I do with + or - in exponencial numbers like:

This says

2^x  -(2^1)/(2^x) - 1 = 0

So ....multiply through by 2^x  =

2^(2x) -2 - 2^x = 0  rearrange

2^(2x) - 2^x  - 2 = 0 

This is a little hard to solve, but we can see that x = 1 is a solution  ... [2^(2*1) - 2*(1) - 2] = [4 - 2 - 2 = 0]

Here's the graph from Desmos.....it indicates that this is the only "real" solution.....

https://www.desmos.com/calculator/v0raxp8fvh

We could take Chris's development a little further as follows:

Given 2^2x - 2^x - 2 = 0 we can rewrite it as (2^x)² - 2^x - 2 = 0

Let y = 2^x, so the last equation can be written as:  y² - y - 2 = 0 or (y + 1)(y - 2) = 0

so y = -1 and y = 2

However, replacing y by 2^x we see that there is no solution to 2^x = -1, so we must have just 2^x = 2, or 2^x = 2¹ giving x = 1 as the only real solution.

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