How do you solve for x: in $2^{x} + 2^{- x} = \frac{5}{2}$ $2^x + 2^-x = 5/2$ ?

Question

2121 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-14T20:50:30

$x = \pm 1$ $x=+-1$

(I'm hoping that someone will come up with a better method of solving, since the following is simply "solution by observation").

We are told: $2^{x} + 2^{- x} = \frac{5}{2}$ $2^x+2^(-x)=5/2$

Observe that
$XXXX \frac{5}{2} = \frac{4}{2} + \frac{1}{2}$ $color(white)("XXXX")5/2 = 4/2 + 1/2$

$XXXXX = 2 + \frac{1}{2}$ $color(white)("XXXXX")=2 + 1/2$

$XXXXX = 2^{1} + \frac{1}{2^{1}}$ $color(white)("XXXXX")=2^1+1/2^1$

$XXXXX = 2^{1} + 2^{- 1}$ $color(white)("XXXXX")=2^1+2^(-1)$

So $2^{x} + 2^{- x} = 2^{1} + 2^{- 1} = 2^{- 1} + 2^{1}$ $2^x+2^(-x)=2^1+2^(-1) =2^-1+2^1$
which implies

$XXX x = \pm 1$ $color(white)("XXX")x=+-1$

Graph

Another possible way is looking at a graph

Graph for #2^x+2^(-x)#

How do you solve for x: in 2x+2−x=522^x + 2^-x = 5/2?