I can't solve (2^x) - 2x = 0 by hand

[Magawu]

I know that the solution is {1,2} but only because the equation is easily solved graphically.
How should I approach solving this using only a simple calculator?

 

[Deleted member 4993]

I know that the solution is {1,2} but only because the equation is easily solved graphically.
How should I approach solving this using only a simple calculator?

Do you know Newton-Raphson's method for estimating roots of Non-Linear functions?

 

[Otis]

How [could] I approach solving [2^x - 2x = 0] using only a simple calculator?

Hi Magawu.

One way is Guess-and-Check (otherwise known as 'The Brute-Force Method'). Subhotosh suggested one form of that. Here's another, for using a simple calculator.

Make a guess: x=3

Now check, using your simple calculator:

2×2×2 - 2×3 = 2

Hmm, the result is too big. Try a smaller guess: x=1.5

…and so on.

EDIT: Solving equations using a simple calculator to guess and check could be a lot of work. Once you find a solution, you need to consider the possibility that there's more than one solution. Are you not allowed to use a graphing/programmable calculator? Otherwise, it might be less work to learn the steps for using a root-finding method (like the one Subhotosh suggested).

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[LCKurtz]

This reply might not fulfill the requirement of "no calculator", depending on your point of view, but you may (or may not) find it interesting, depending on your background. Your problem is similar to asking to solve [imath]xe^x =y[/imath] for [imath]x[/imath]. Since this is just the problem of finding the inverse function, it has been studied and the inverse function is called the Lambert W function so you have [imath]x=W(y)[/imath], and many advanced mathematics calculation packages include this function just like they include trig functions and exponentials etc. You can use it algebraically solve your problem and other similar problems. You are starting with [imath]2^x=2x[/imath] and you want to solve for [imath]x[/imath]. The idea is to get it in the form of the equation above to use the Lambert W function to solve it. Follow these steps showing the algebra to get your problem in the correct form:
[math]2^x=2x\\ e^{x\ln 2}=2x\\ 1=2xe^{-x\ln 2}\\ \frac 1 2=xe^{-x\ln 2}\\ -\frac 1 2\ln 2=-x\ln 2 e^{-x\ln 2}[/math]This is in the required form so we can write
[math]-x\ln 2 = W(-\frac 1 {2\ln 2})\\ x = -\frac 1 {\ln 2}W(-\frac1 2 \ln 2)[/math]and you have solved your problem for [imath]x[/imath]. Now, if you plug that expression in a calculator you will get [imath]x=1[/imath].
Of course, this is a lot more work than just observing the answer is [imath]x=1[/imath], but that isn't the point. Just like in some problem where you might get an answer with pencil and paper of [math]x=\sin(\frac {5\pi} 8)[/math] you could either leave it that way or calculate a decimal, in your problem you could leave it in terms of the W function or calculate a decimal answer.

 

[Steven G]

Hi Magawu.

One way is Guess-and-Check (otherwise known as 'The Brute-Force Method'). Subhotosh suggested one way. Here's another, for using a simple calculator.

Make a guess: x=3

Now check, using your simple calculator:

2×2×2 - 2×3 = 18

Hmm, the result is too big. Try a smaller guess: x=2

…and so on. Questions?

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2×2×2 - 2×3 = 18???

 

[Otis]

2×2×2 - 2×3 = 18???

Thanks, Jomo. I'd started with x=4 and then changed my mind and botched the edit. I will go to the corner for 18 minutes.

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