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A Simple, Yet Difficult, Math Problem

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I recently came across a math problem I cannot solve, though it looks solvable. This is it:

3^x + 4^x = 5^x

As you can see, you instantly know the answer is 2(Pythagoras theorem), but how do you go about solving it?

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What do you get when you substitute numbers for "x?" Do any of them give a solution that work?

If x=0:

3^0 + 4^0 = 5^0

1 + 1 = 1

NOPE.

If x=1:

3^1 + 4^1 = 5^1

3 + 4 = 5

NOPE.

If x=2:

3^2 + 4^2 = 5^2

9 + 16 = 25

DING!!

;)

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According to Fermat's Last Theorem, 2 is the highest integer power for any such equation....

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Integer values for x weren't specified in post 1.

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I know that. I was simply pointing out a related topic to the problem...

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I recently came across a math problem I cannot solve, though it looks solvable. This is it: 3^x + 4^x = 5^x

...but how do you go about solving it?

You could split it into two independent (simultaneous) equations:

y = 3^x + 4^x

y = 5^x

Then, lessee here, take the Log of each equation:

R = Log y = x / Log 3 + x / Log 4

R = x / Log 5

...I think. :D

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And it took a while to prove Fermat's too.

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What do you get when you substitute numbers for "x?" Do any of them give a solution that work?

If x=0:

3^0 + 4^0 = 5^0

1 + 1 = 1

NOPE.

If x=1:

3^1 + 4^1 = 5^1

3 + 4 = 5

NOPE.

If x=2:

3^2 + 4^2 = 5^2

9 + 16 = 25

DING!!

;)

Can you do it by any other means other than trial and error? I mean what if the problem was actually this:

4^x + 5^x = 6^x

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I'd say 99.9% it has to be done iteratively.

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4^x+5^x=6^x, x=2.487939 :weather_storm:

I'll get back to ya later, they're calling me to dinner.

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Imagine one of the x's in the equation as being the only x and the others are constants. Solve for that x. Then you get a forumla like x=f(x). Split it up into two forumas y=x and y=f(x). If you plot the two formulas on an x-y graph the point where they intercept is going to be the x you're looking for in the initial equation.

Lot of times you can solve this iteratively. You make a guess for x that's roughly the intercept point, then you put it into the formula y=f(x). The value you calculate for f(x) becomes the new x. Keep doing that until hopefully it converges to so many decimal places.

If it doesn't converge you can try sovling for another x in the initial equation. Or you can change y=f(x) into y=(x+f(x))/2, or y=(2*x+f(x))/3, etc. It's like you're flattening out f(x) towards the line y=x. The slope of f(x), f'(x) at the intercept of the line y=x can't be too large or the iterative x's spiral outward intstead of inward toward the real solution. If there's no solution and you plot out y=x and f(x) they won't ever cross paths. Unless you extend the range into complex numbers. Then you get all kinds of interesting stuff. Anyway,

3^x+4^x=5^X

log(3^x+4^x)=log(5^X)

=X log(5)

X=log(3^x+4^x)/log(5)=f(x)

guess x=1.9

1.9 1.920359 1.936563 1.949464 1.959738 1.967921 1.974439 1.979631

1.983768 1.987065 1.989691 1.991785 1.993453 1.994782 1.995841 1.996685

1.997358 1.997894 1.998322 1.998662 1.998934 1.99915 1.999323 1.99946

1.99957 1.999657 1.999727 1.999782 1.999826 1.999861 1.999889 1.999912

1.99993 1.999944 1.999955 1.999964 1.999971 1.999977 1.999982 1.999985

1.999988 1.99999 1.999992 1.999994 1.999995 1.999996 1.999997 1.999997

1.999998 1.999998 1.999998 1.999999 1.999999 1.999999 1.999999 1.999999

guess x=2.1

2.1 2.079759 2.063606 2.050718 2.040438 2.032239 2.025701 2.020487

2.016331 2.013017 2.010376 2.00827 2.006592 2.005254 2.004187 2.003337

2.00266 2.00212 2.001689 2.001346 2.001073 2.000855 2.000681 2.000543

2.000433 2.000345 2.000275 2.000219 2.000174 2.000139 2.000111 2.000088

2.00007 2.000056 2.000044 2.000035 2.000028 2.000022 2.000018 2.000014

2.000011 2.000009 2.000007 2.000005 2.000004 2.000003 2.000003 2.000002

2.000001 2.000001 2.000001 2 2 2 2 *

3 and 4 are fairly close to each other so you can find a good first guess by

approximating 3^x+4^x=5^x to 2*[(3+4)/2]^x=5^x

2*[(3+4)/2]^x=5^x

2=[2*5/(3+4)]^x=[10/7]^x

x=log2/log(10/7)=log2/[log10-log7]

x=1.943358210

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Or, you guys could recognize another property of 3, 4, and 5. They make up the sides of a right triangle. Knowing that, the question becomes simple.

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Obviously Dave, Sebastian even said so in #1, the puzzle is to do it "without knowing that" by just treating it as an equation.

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• 4 weeks later...

While we're on the topic of simple problems, I'll throw one in...

a + b = -3 and ab = 4, then a^3 + b^3 equal...

enjoy!

a^3+b^3=9

thanks :)

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Correct me if I'm wrong but is there a mistake in that problem. Should it be a+b=-3, ab=-4 and a^3+b^3=? ?

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That's one yes for the first post after me and a no for the last one. You can't solve it like a normal algebraic system.

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