2^x = x^2 Is there anyway to find X other than drawing the graph?

Question

and other than trial and error

2
originally it asked to solve a^b = b^a. and a

Answer 1

Hi,

This one needs to draw the graph as Y1 = 2^x - x^2 where you can find the value of its x intercept as your x value. You could also let Y1 = 2^x and Y2 = x^2 and find the x value of their intersection. Either way you solve, the x values where that occurs should be the same, at x = -.7666647 or x = 2 or x = 4.

I hope that helps!! :-)

Answer 2

All the "x = 4" answers are wrong.

There are two solutions, and you do not need logs to get them! First, you can see by inspection that x=0 is a solution because then you have zero on both sides.

To look for nonzero solutions, simply divide both sides by x. Then the answer drops right out: the left side just becomes "2" and the right side just "x", so your other solution is x=2.

In short, two solutions: x=0, and x=2.

If you wanted to do it graphically, draw two curves: y=2x and y=x^2. The first one is a straight line passing through (0,0), with a slope of 2. The second is a parabola whose bottom touches (0,0). The line crosses the parabola in two places: at the origin (the x=0 solution) and at the point (2,4) (thus x=2).

Answer 3

Try taking logs of both sides
x*ln(2) = 2*ln(x)

We know x =2 is a solution.
But are there any other solutions?

At the point x = 2
2*ln(x) has gradient 2/x = 2/2 = 1
The line x*ln(2) has gradient ln(2) > 1

Since the two curves do not have the same gradient at the point of intersection, and we know that the gradient of the ln(x) graph is decreasing, there must be another point of intersection.

You need to find that numerically using the Newton Raphson method.

Let f(x) = x*ln(2) - 2*ln(x)
f'(x) = ln(2) - 2/x

So x_new = x_0ld - f(x_old)/f'(x_old)
= x_old - [x_old*ln(2) - 2*ln(x_old)] / [ln(2) - 2/x_old]

Start with x_old = 4 or something and iterate.
----------------------------
Actually it turns out that x = 4 is an exact solution.

So you two answers are x = 2 and x = 4.

Answer 4

there are 2 cases
1) x is integer
2^x = x^2

by inspection x = 2 is a solution as 2^2 = 2^2 is identity

let us look for other solutions

x has to be power 2.

(reason if x has a factor other than 2 then RHS has a factor other than 2 but LHS does not)

so x> 2( x= 2 is considered)

however we shall take x = 2y to make our equation simpler

2^(2y) = (2y)^2
(2^y)^2 = (2y)^2
2^y = 2y
2^(y-1) = y

now RHS is linear and LHS is exponentional and y = 3 makes

2^2 = 4 > 3
so y < 3
y= 2 makes 2^1 = 2 correct
y = 1 makes 2^0 = 1 correct
y =2 gives x = 4

and y = 1 gives x = 2 (already found)
so x = 2 and 4 are 2 solutions

This is more complex for rational solution

Answer 5

The integral solutions are 2 and 4. 0 is not a solution; 2^0 equals 1, but 0^2 equals 0.

Answer 6

Well, your equation gives
x log 2 = 2 log x,
or x/log x = 2/log 2
Obviously x = 2 is a solution and so is x = 4.
To find the remaining solutions I would
get an approximate solution and then use
Newton's method to hone in on the answer.
Also, if x>4 there are no more solutions,
since x/log x is an increasing function for x > 4.

Answer 7

Trial.

Try x = 2.

2^(2) = 4 = (2)^2.

Answer 8

y graph?? not needed
c look at it
2^x=x^2
thats it!
x=2
and c for another possible value
2^4 is 16 and 4^2 is also 16
so the answer can also be x=4

Answer 9

Just look at it, it just switches the terms so it is 2.

Answer 10

if x=2 you can get it by inspection.