Question

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

Answer 1

`x = log_2(5/2+sqrt(29)/2)~~2.37645`

Explanation:

We have:

`2^x-2^-x=5`

Take the `5` over to the other side to get:
`2^x-5-2^-x=0`

Now multiply the whole thing through by `2^x` and we get:

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

We substitute `t=x^2` and that gives us:

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

We now have a quadratic that can be solved using the quadratic formula with terms `a=1, b=5` and `c=-1`.

`-> t = (5 +- sqrt(5^2-4*1*(-1)))/(2*1)=(5+-sqrt(29))/2`

Thus `t= 5/2+sqrt(29)/2` or `t=5/2-sqrt(29)/2`

Remeber `2^x=t`. Reverse the substitution of `t` to obtain:

`2^x=5/2+sqrt(29)/2`

Now take the logarithm to the base `2` and we get:

`x = log_2(5/2+sqrt(29)/2)~~2.37645`

We can ignore the `t=5/2-sqrt(29)/2` as it would involve taking the logarithm of a negative number which we wish to avoid.