Sketching the graphs, there are three solutions:
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 for which the following equation would be true.
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?
The solutions of an equation of the form 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.
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.
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.
---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.