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anonymous
 one year ago
Calculus help needed! Will medal and fan!!
anonymous
 one year ago
Calculus help needed! Will medal and fan!!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Where f(x)=x^2 and g(x)=2^x, I am looking for the points of intersection. So far, I have started this by setting 2^x=x^2 to look for x. I worked it to xlog2(2)=2log2(x) and then x(1)=2log2(x) and then x/2=log2(x) and then x/2=ln(x)/ln(2) but now feel as if I am working in a circle... eventually, I reduced it to x=e^(xln(2)/2), however, this still has x on both sides. Help, please?

Plasmataco
 one year ago
Best ResponseYou've already chosen the best response.0im only 13 but sure, I can try.

Plasmataco
 one year ago
Best ResponseYou've already chosen the best response.0http://media.education2020.com/contentengine/tools/Calculators/graphingcalculator.html

Plasmataco
 one year ago
Best ResponseYou've already chosen the best response.0type in your equations and get the answer!

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.2Looks like you will need the Product Log Function  https://en.wikipedia.org/wiki/Lambert_W_function#Example_1 http://www.wolframalpha.com/input/?i=x%5E2%3D2%5Ex

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.2$$ x^2=2^x\\ 2\ln x=x\ln2\\ \cfrac{\ln x}{x}=\cfrac{\ln2}{2}\\ \ln x \exp^{\ln x}=\cfrac{\ln2}{2}\\ \ln x \exp^{\ln x}=\cfrac{\ln2}{2}=W\left(\cfrac{\ln 2}{2}\right)\\ \ln x=W\left(\cfrac{\ln2}{2}\right)\\ x=\exp^{W\left(\cfrac{\ln2}{2}\right)}\\ =\frac{2W\left(\cfrac{\ln2 }{2}\right)}{\ln 2}=2 $$ Where \(W()\) is the Lambert W function  https://en.wikipedia.org/wiki/Lambert_W_function Another solution is, since x can be positive or negative: $$ =\frac{2W\left(\cfrac{\ln2 }{2}\right)}{\ln 2}\approx0.77 $$ Now check if 2^2 is another solution by plugging in to original equation. If so, that will be a third solution.
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