Need a MATH WHIZ

[GoldenBear]

How would one solve this out the amish way..

x^2=2^x

The answers are x=2 and x=4..and another number I think.

No guess and check, or graphing on a calc..just the old fashioned way. You can use logs on a calc of course, but otherwise..just some old-school math.

I don't know if it's even possible, but it MUST be right?

 

[dopcombo]

well if u just do a simple sketch of the 2 curves u get 2 curves that intersect each other at 3 points.
x =2, 4, as well as a negative value.

2^x is exponential so its gradient increases faster than x^2.

meanwhile, u've got me intrigued
i dun believe i can;t solve this.

 

[Mister T]

is it not possible to solve for explicitly

 

[eakers]

id take the ln of both sides to get at the exponent
then use e^ln of both sides
but i dunno if thats right
something like that though.

*kat. <-- could be wayy off.

 

[BowDown]

If my graphing calc wasn't in the car I could help ya .

 

[Unsickle]

using log props you can separate the X to one side.

you get

log(x)/x = log(2)/2

That's the most simplified form.

 

[dopcombo]

which then gives 2 as a trivial answer
but how to get the rest is not obvious from that unless u trial and error (which is prohibited)

 

[Scrapster]

using log props you can separate the X to one side.

Care to share those log props?

 

[dopcombo]

x^2 = 2^x

log x^2 = log 2^x
<1 of property of logs>
2 log x = x log 2

(log x)/x = (log 2)/2

 

[GoldenBear]

So far I have..

x^2 = 2^x
2log x = x log2
log x/x = log 2/2 (divide one side by x and the other by 2)

So I don't think you can get farther..

D'oh Dop beat me to it.

 

[Unsickle]

it's a transcendental equation. Nothing comes to mind that would solve it for all possible solutions unless you turned it somehow into a third order polynomial. Shrug.

 

[Noriaki]

Scrapster:

x^2 = 2^x

take the log of both sides (any base you want, most people choose natural though (log base e)).

ln (x^2) = ln (2^x)

by the log property ln (x^y) = y*ln(x) you get

2*ln(x) = x*ln(2)

cross division

ln(x)/x = ln(2)/2


But as dopcombo pointed out this doesn't help much, but to give you the trivial solution of x = 2, there is x = 4, some x < 0 as well.

 

[josphstalinator]

use taylor series to expand log x then solve the polynomial

 

[dopcombo]

man, how can i forget taylor series

yeah i guess that is the closest u can come to using algebra since logarithms generally dun work very well with algebraic manipulations.

well one other way u could do it would be to use the sketches and guess 3 initial points, use some sort of iteration method (newton-raphson comes to mind) and approximate towards the answer.

my math major ICQ friend just confirmed that it is a transcendental eqn which means that it cannot be solved by algebra. You gotta plot it or use a computer to solve it.

damn, i'm bored

 

[Scrapster]

Is anyone here really good with Taylor series? We just got done going over it briefly in class. I'm curious on how we could find the points it converges too.

Where's arod?

 

[Noriaki]

Damn it's only been like 12 months since calculus class and I forgot Taylor series already...damn...

 

[Scrapster]

I know the Taylor series definitions for ln(x)@x=1 and ln(1+x)@x=0 if that can jump start anyone.

ln(1+x) @ x=0

Summation ((-1)^(n-1))*(x^n) all over n
n=1 - infinity

ln(x) @ x=1 is just the same thing except it's (x-1)^n

 

[erub]

we did this problem in class about a month ago (precal)...it turns out that there are 4 solutions I think - the 2 that you named, and 2 unknowns...

 

[arod324]

Right now, we would have to find out an equation that describes the sum of all of the parts... and then we need to sum them up.... give me a sec to do some calculations.

 

[Scrapster]

I get 2, 4, and -.765 on my graph.

 

[SnoopCat]

hypotically, it is imperative to understand the the negative roots for x^2 can only be determined in the 3rd dimensional quadrant. In theory, you must use quantum anaylsis with interogattive mathematicates to approach this problem discretly. Find the summation from the imaginary root i to the lateral induction method by the use of indirect proof.

 

[Mday]

it's a very easy problem ;-)

 

[Turkey]

[SnoopCat puts down the joint]

I told you it doesn't make you stupider! Now try this one...

 

[Mday]

in the R^2 (euclidean space, 2 dimensions, reals), there are only 2 answers for x > 0

first of all, knowing the roots of x^2 does not help, since that means one side is 0, and as we all know 2^x is never zero except for when x is negative infinite.

2^x has an increasing slope for x > 0, as does x^2, for x>0, so there can be at most 2 intersections. else there exists some point (rather interval, since both functions are continuous) where at least one of the functions have a decreasing slope.

however, there is another value when x is negative ;-)

 

[Scrapster]

Mday, can you solve this with Taylor series?