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

$${{\mathtt{2}}}^{{\mathtt{x}}}{\mathtt{\,-\,}}{{\mathtt{2}}}^{\left({\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{x}}\right)}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0}}$$

Tiagunix
Feb 18, 2015

#1**+10 **

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

CPhill
Feb 18, 2015

#1**+10 **

Best Answer

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

CPhill
Feb 18, 2015

#2**+5 **

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: y^{2} - 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} = 2^{1 }giving x = 1 as the only real solution.

.

Alan
Feb 20, 2015