Question

How do you solve `2^x + 2^(1/x) = 18` ?

Answer 1

This equation has no closed form solution.

We can calculate approximations for the solutions using Newton's method:

`x_1 ~~ 4.0716264696514`

`x_2 ~~ 0.2456020972095`

Explanation:

This is an interesting question because of the similar-looking but much easier to solve equation:

`2^x * 2^(1/x) = 18`

This has a closed form solution in terms of `log`s.

`2^x*2^(1/x) = 2^(x+1/x)`

So we get:

`(x+1/x) log(2) = log(2^(x+1/x)) = log(18)`

Hence a quadratic in `x`:

`log(2)x^2-log(18)x+log(2) = 0`

Hence solutions:

`x = (log(18)+-sqrt(log(18)^2-4log(2)^2))/(2log(2))`

Our current problem has no such solution in closed form in terms of ordinary functions.

We can find a numeric approximation using Newton's method.

Let `f(x) = 2^x + 2^(1/x) - 18`

Then `f'(x) = ln(2) (2^x - 2^(1/x)/(x^2))`

We find:

`f(4) = 2^4+2^(1/4)-18 ~~ -0.8`

So let's use that as our first approximation `a_0 = 4` and iterate using the formula:

`a_(i+1) = a_i - f(a_i)/(f'(a_i)) = a_i - (2^x+2^(1/x)-18)/(ln(2) (2^x - 2^(1/x)/(x^2)))`

Evaluating this in a spreadsheet I got successive approximations:

`a_1 = 4.07344912583206`

`a_2 = 4.0716276291883`

`a_3 = 4.07162646965187`

`a_4 = 4.07162646965141`

The other root will be about `1/a_4 ~~ 0.2456020972094757`