How do you solve $2^(x) - 2^(-x) = 5$ ?

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Answer 1 · 2016-07-08T17:56:06

$x = log_2(5+sqrt(29))-1 ~~ 2.3764522$

Let $t = 2^x$

Then the equation becomes:

$t-1/t = 5$

Multiply through by $4t$ and rearrange slightly to get:

$0 = 4t^2-20t-4$

$= (2t-5)^2-29$

$= (2t-5)^2-(sqrt(29))^2$

$=((2t-5)-sqrt(29))((2t-5)+sqrt(29))$

$=(2t-5-sqrt(29))(2t-5+sqrt(29))$

Hence:

$2^(x+1) = 2*2^x = 2t = 5+-sqrt(29)$

For any Real value of $a$ , $2^a > 0$ .

So (for Real solutions) we require:

$2^(x+1) = 5+sqrt(29)$

Hence:

$x+1 = log_2(5+sqrt(29))$

So:

$x = log_2(5+sqrt(29))-1$

To calculate a numerical value for $log_2(5+sqrt(29))$ we can use the change of base formula to find:

$log_2(5+sqrt(29)) = ln(5+sqrt(29))/ln(2) ~~ 3.3764522$

So:

$x ~~ 3.3764522 - 1 = 2.3764522$

How do you solve 2^(x) - 2^(-x) = 52^(x) - 2^(-x) = 5?