Richard Z.

asked • 11/24/14# Can 2^x = x^2 be solved explicitly? I can see that {2,4} are solutions, but the calculator shows a third solution, x=-0.7666647.

2

^{x}= x^{2}How would you solve this?

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## 2 Answers By Expert Tutors

Francisco P. answered • 11/24/14

Tutor

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(297)
Rigorous Physics Tutoring

If you are taking or have taken calculus, you could follow the following:

We can solve for it numerically using Newton's method. Let f(x) = 2

^{x}- x^{2}. Then f'(x) = ln2 2^{x}- 2x.x

_{n+1}= x_{n}- f(x_{n})/f'(x_{n}) is used iteratively to find the root of f(x) given a starting value close to the desired root.From above,

x

_{n+1 }= x_{n }- (2^{x}- x^{2})/(ln2 2^{x }- 2x)Let's start with x

_{0}= -0.5.x

_{1}= -0.5 - (2^{-0.5}- (-0.5)^{2})/(ln2 2^{-0.5 }- 2(-0.5)) = -0.80675650...Use this value of x

_{1}to find x_{2}and so on until the iteration doesn't change in the next iteration. A graphing calculator with an ANS and ENTRY key would work very nicely.x

_{2}= -0.76735361...x

_{3 }= -0.76666491...x

_{4}= -0.76666469...x

_{5 }= -0.76666469...So, it looks like -0.76666469.... is a solution to 2

^{x }= x^{2}.Francisco P.

Are there any other methods besides graphing that you know of, Christopher?

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11/24/14

Christopher R.

tutor

Perhaps, your right. This would be a calculus problem if you had to solve this problem other than graphing. However, one can setup this problem as a function being f(x)=2

^{x }- x^{2 }and make a table containing intelligent guesses to find what values of x would make the function equal to zero. Then use interpolation to find the approximate value of x. For example, let x=0 in which makes f(0)=1 and x=-1 in which makes f(-1)=1/2-1=-1/2. I think you can see what I'm saying about how one could use interpolation to find some approximate value of x being in between x=0 and x=-1 to make the original equation to be true.Moreover, this could be one way to show the student on how to do the problem if he is taking Algebra II.

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11/24/14

Francisco P.

I didn't think of that. Thanks and happy tutoring.

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11/25/14

Christopher R.

tutor

Your welcome, you could see that I didn't initially think of that either... Lol. Thanks for your answer and happy tutoring to you also.

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11/25/14

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Christopher R.

11/24/14