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

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

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Answer 1 · 2016-07-08T17:56:06

$x = {log}_{2} (5 + \sqrt{29}) - 1 \approx 2.3764522$ $x = log_2(5+sqrt(29))-1 ~~ 2.3764522$

Explanation:

Let $t = 2^{x}$ $t = 2^x$

Then the equation becomes:

$t - \frac{1}{t} = 5$ $t-1/t = 5$

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

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

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

$= {(2 t - 5)}^{2} - {(\sqrt{29})}^{2}$ $= (2t-5)^2-(sqrt(29))^2$

$= ((2 t - 5) - \sqrt{29}) ((2 t - 5) + \sqrt{29})$ $=((2t-5)-sqrt(29))((2t-5)+sqrt(29))$

$= (2 t - 5 - \sqrt{29}) (2 t - 5 + \sqrt{29})$ $=(2t-5-sqrt(29))(2t-5+sqrt(29))$

Hence:

$2^{x + 1} = 2 \cdot 2^{x} = 2 t = 5 \pm \sqrt{29}$ $2^(x+1) = 2*2^x = 2t = 5+-sqrt(29)$

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

So (for Real solutions) we require:

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

Hence:

$x + 1 = {log}_{2} (5 + \sqrt{29})$ $x+1 = log_2(5+sqrt(29))$

So:

$x = {log}_{2} (5 + \sqrt{29}) - 1$ $x = log_2(5+sqrt(29))-1$

To calculate a numerical value for ${log}_{2} (5 + \sqrt{29})$ $log_2(5+sqrt(29))$ we can use the change of base formula to find:

${log}_{2} (5 + \sqrt{29}) = \frac{ln (5 + \sqrt{29})}{ln (2)} \approx 3.3764522$ $log_2(5+sqrt(29)) = ln(5+sqrt(29))/ln(2) ~~ 3.3764522$

So:

$x \approx 3.3764522 - 1 = 2.3764522$ $x ~~ 3.3764522 - 1 = 2.3764522$

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