# Thread: Exponential Problem (Quite hard)

1. ## Exponential Problem (Quite hard)

Here's a problem i really can't solve. I've tried logs and spent 2 science periods and an accounting period on it.

Find values of $x$ for which the following equation would be true.

$x^2 = 2^x$

By inspection we can see x = 2 but i'm sure this isn't the only solution.

Even my teacher can't solve this one. Am i missing something simple here?

2. Sketching the graphs, there are three solutions:
$x_1\in(-1,0),x_2=2,x_3=4$

3. Originally Posted by red_dog
Sketching the graphs, there are three solutions:
$x_1\in(-1,0),x_2=2,x_3=4$
Thank you Red_Dog, but i wish to see an algebraic solution please. I want to see how it should be solved.

And please explain what you mean with this: $x_1\in(-1,0)$

4. The solutions of an equation of the form $f(x)=g(x)$ are the x-coordinates of the points of intersection between the function's graphs.
In the sketch, a point of intersection has the x-coordinate between -1 and 0. So, a solution of the equation is in that interval.

5. Originally Posted by red_dog
The solutions of an equation of the form $f(x)=g(x)$ are the x-coordinates of the points of intersection between the function's graphs.
In the sketch, a point of intersection has the x-coordinate between -1 and 0. So, a solution of the equation is in that interval.
But i just cannot see a negative value satisfying the equation...

6. Probably you can't find an exact negative solution, but this solution exists.
Let $f(x)=x^2-2^x$
We have $f(-1)=1-\frac{1}{2}=\frac{1}{2}>0$ and $f(0)=-1<0$.
$f$ is continue and it has the Darboux property. So $f$ must have at least a solution in $(-1,0)$.

7. Originally Posted by red_dog
Probably you can't find an exact negative solution, but this solution exists.
Let $f(x)=x^2-2^x$
We have $f(-1)=1-\frac{1}{2}=\frac{1}{2}>0$ and $f(0)=-1<0$.
$f$ is continue and it has the Darboux property. So $f$ must have at least a solution in $(-1,0)$.
Is there no way of taking a pen and paper and working this out? All the solutions it has.

8. Originally Posted by janvdl
Here's a problem i really can't solve. I've tried logs and spent 2 science periods and an accounting period on it.

Find values of $x$ for which the following equation would be true.

$x^2 = 2^x$

By inspection we can see x = 2 but i'm sure this isn't the only solution.

Even my teacher can't solve this one. Am i missing something simple here?
Yeah, sometimes you encounter such equations that can't be solved the "normal" way. I think there's a name for such equations.

They can be solved graphically, as shown.

And they can be solved analytically, or by "pen and paper" or calculator. It is by trial-and-error, by repetition, by iteration.

Before I learned about the Newton's Method or Newton-Rhapson Method, I was already using iteration when I cannot solve quadratic/cubic/etc/trigo equations. . Newton's Method made it easier for me because it is systematic and after only about 3 iterations, depending on how close your seed root/factor is, you'd almost get the real root/factor.

So,
x^2 = 2^x
x^2 -2^x = 0
Let y = x^2 -2^x --------(i)
Use Newton's Method of iteration on (i). If x would make y zero or very close to zero, then you got your x. There can be many of those x's, though.

x2 = x1 -[y1 / (y1')] -------Newton's Method.
where
---x1 is the seed x. The x you assume to be a root or factor.
---x2 is the next x that should be closer to the correct x than x1
---x3 is the next x that should be closer to the correct x than x2
---x4 is the next x that should be closer to the correct x than x3
...(iteration, you know. I usually stop at x3 or x4.)
---y1 is value of y when x = x1
---(y1)' is value of first derivative of y when x = x1.

9. I get $x \approx -0.7666647$.

-Dan

10. There are no general way to solve equations. The only thing we can show is that a solution exists.