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avatar+11 

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

Best Answer 

 #1
avatar+85809 
+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
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2+0 Answers

 #1
avatar+85809 
+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
avatar+26640 
+5

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

 

Given 22x - 2x - 2 = 0 we can rewrite it as (2x)2 - 2x - 2 = 0

 

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

.

Alan  Feb 20, 2015

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