Question

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

Answer 1

`x=log_2(5+sqrt29)-1`

Explanation:

Since `a^-b=1/a^b`, the equation can be rewritten as

`2^x-1/2^x=5`

Multiply each term by `2^x`.

`2^x(2^x)-1/2^x(2^x)=5(2^x)`

To simplify `2^x(2^x)`, use the rule `a^b(a^c)=a^(b+c)`:

`2^(x+x)-1=5(2^x)`

`2^(2x)-5(2^x)-1=0`

Let `u=2^x`. Then, `2^(2x)=(2^x)^2=u^2` and `5(2^x)=5u`.

`u^2-5u-1=0`

Use the quadratic formula to show that

`u=(-(-5)+-sqrt(5^2-(4*-1*1)))/(2*1)=(5+-sqrt29)/2`

Recall that `u=2^x`.

`2^x=(5+sqrt29)/2`

`x=log_2((5+sqrt29)/2)`

Using `log_a(b/c)=log_a(b)-log_a(c)`:

`x=log_2(5+sqrt29)-log_2(2)`

`x=log_2(5+sqrt29)-1`

As for the negative version of `(5+-sqrt29)/2`, we see that

`2^x=(5-sqrt29)/2`

Has no solutions since `(5-sqrt29)/2<0` and `2^x>0` for all values of `x`.

Answer 2

`x = log_e(1/2 (5 + sqrt[29]))/log_e(2)`

Explanation:

Taking `y = 2^x` we have

`y -1/y=5->y^2-5y-1=0`

Solving for `y`

`y = 1/2 (5 - sqrt[29])`, `y = 1/2 (5 + sqrt[29])`

but `1/2 (5 - sqrt[29]) < 0` and `y = 2^x > 0` Now solving for `x`

so

`x = log_e(1/2 (5 + sqrt[29]))/log_e(2)`