Question

Solve this question and answer elaborately as soon as possible. Fast... Plzzz...

Answer 1

x x^2 2^x
-2 4 0.25
-1 1 0.5
0 0 1
1 1 2
2 4 4
3 9 8
4 16 16
5 25 32
6 36 64
10 100 1024

So as you can see, they are both equal when x = 2 or x = 4.

Also notice that before both of them are equal, one is bigger then the other and after they switch. For example,
at x = 1, 2^x > x^2
x = 2, 2^x = x^2
but at x = 3, x^2 > 2^x

They seem to switch at x = 2, x = 4 and somewhere between x = -1 and 0 so were going to have another answer for x there. The best step to take from here would be to move both terms to one side to get a new equation
2^x = x^2
2^x - x^2 = 0
f(x) = 2^x - x^2

At this point were going to need to use the The Newtons method. The formula for it is:
x[i+1] = x[i] - f(x[i]) / f'(x[i])
This formula approximates values for x that will make f(x) = 0. Each time you computer it you will get close to a value of x that will make f(x) = 0.

So we need to take the derivative of f(x)
f'(x) = 2^x * ln(2) - 2x
and plug these both into our formula,
x[i+1] = x[i] - (2^x[i] - x[i]^2) / (2^x[i] * ln(2) - 2x[i])

Now we need to pick a point set our x[0] to, our starting point. Since we know the point we are looking for is between 0 and 1 well start with 0. (If we were to start with 1, our formula would go towards 2 because that is the closest value that makes f(x) = 0).

x[0] = 0
x[1] = 0 - (2^0 - 0) / (2^0 * ln(2) - 0) = -1 / (1 * ln(2)) = -1 / ln(2) = -1.442...
x[2] = -1.442 - (2^(-1.442...) - -1.442...^2) / (2^(-1.442...) * ln(2) - 2 * -1.442...) = -0.897
...
x[10] = -0.7666

So we have x = 2, 4, and about -0.7666