Question

Find the solutions of `x^2=2^x`?

Answer 1

`x_1 = -0.7667` to 4dp
`x_2= 2`
`x_3 = 4`

Explanation:

`x^2 = 2^x`

There is no "easy" way to solve this equation.

First let us look at the graphs:

We can see there is one solution in the interval `-1 < x < 0` and a second solution at what appears to be `x=2`, and a third at what appears to be `x=4`. The two solution `x=2,4` can easily be verified to be exact by substitution.

We can find the third solution numerically, using Newton-Rhapson method

Let `f(x) = x^2-2^x => f'(x) = 2x-(ln2)2^x`, and using the Newton-Rhapson method we use the following iterative sequence

`{ (x_0=-1), ( x_(n+1)=x_n - f(x_n)/(f'(x_n)) ) :}`

Then using excel we can tabulate the iterations as follows:

And we conclude that the remaining solution is `x=-0.7667` to 4dp