Sebastian Posted December 13, 2006 Report Share Posted December 13, 2006 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? TheBigDog 1 Quote Link to comment Share on other sites More sharing options...

InfiniteNow Posted December 13, 2006 Report Share Posted December 13, 2006 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 = 5^01 + 1 = 1NOPE. If x=1:3^1 + 4^1 = 5^13 + 4 = 5NOPE. If x=2:3^2 + 4^2 = 5^29 + 16 = 25DING!! ;) Quote Link to comment Share on other sites More sharing options...

C1ay Posted December 13, 2006 Report Share Posted December 13, 2006 According to Fermat's Last Theorem, 2 is the highest integer power for any such equation.... Quote Link to comment Share on other sites More sharing options...

ughaibu Posted December 13, 2006 Report Share Posted December 13, 2006 Integer values for x weren't specified in post 1. Quote Link to comment Share on other sites More sharing options...

C1ay Posted December 13, 2006 Report Share Posted December 13, 2006 I know that. I was simply pointing out a related topic to the problem... Quote Link to comment Share on other sites More sharing options...

Pyrotex Posted December 13, 2006 Report Share Posted December 13, 2006 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^xy = 5^x Then, lessee here, take the Log of each equation:R = Log y = x / Log 3 + x / Log 4R = x / Log 5 ...I think. :D Quote Link to comment Share on other sites More sharing options...

Qfwfq Posted December 13, 2006 Report Share Posted December 13, 2006 And it took a while to prove Fermat's too. Quote Link to comment Share on other sites More sharing options...

Sebastian Posted December 14, 2006 Author Report Share Posted December 14, 2006 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 = 5^01 + 1 = 1NOPE. If x=1:3^1 + 4^1 = 5^13 + 4 = 5NOPE. If x=2:3^2 + 4^2 = 5^29 + 16 = 25DING!! ;) 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 Quote Link to comment Share on other sites More sharing options...

Qfwfq Posted December 14, 2006 Report Share Posted December 14, 2006 I'd say 99.9% it has to be done iteratively. Quote Link to comment Share on other sites More sharing options...

max4236 Posted December 17, 2006 Report Share Posted December 17, 2006 4^x+5^x=6^x, x=2.487939 :weather_storm: I'll get back to ya later, they're calling me to dinner. Quote Link to comment Share on other sites More sharing options...

max4236 Posted December 18, 2006 Report Share Posted December 18, 2006 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=1.91.9 1.920359 1.936563 1.949464 1.959738 1.967921 1.974439 1.9796311.983768 1.987065 1.989691 1.991785 1.993453 1.994782 1.995841 1.9966851.997358 1.997894 1.998322 1.998662 1.998934 1.99915 1.999323 1.999461.99957 1.999657 1.999727 1.999782 1.999826 1.999861 1.999889 1.9999121.99993 1.999944 1.999955 1.999964 1.999971 1.999977 1.999982 1.9999851.999988 1.99999 1.999992 1.999994 1.999995 1.999996 1.999997 1.9999971.999998 1.999998 1.999998 1.999999 1.999999 1.999999 1.999999 1.999999guess x=2.12.1 2.079759 2.063606 2.050718 2.040438 2.032239 2.025701 2.0204872.016331 2.013017 2.010376 2.00827 2.006592 2.005254 2.004187 2.0033372.00266 2.00212 2.001689 2.001346 2.001073 2.000855 2.000681 2.0005432.000433 2.000345 2.000275 2.000219 2.000174 2.000139 2.000111 2.0000882.00007 2.000056 2.000044 2.000035 2.000028 2.000022 2.000018 2.0000142.000011 2.000009 2.000007 2.000005 2.000004 2.000003 2.000003 2.0000022.000001 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^x2=[2*5/(3+4)]^x=[10/7]^xx=log2/log(10/7)=log2/[log10-log7]x=1.943358210 Quote Link to comment Share on other sites More sharing options...

pgrmdave Posted December 18, 2006 Report Share Posted December 18, 2006 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. Quote Link to comment Share on other sites More sharing options...

Qfwfq Posted December 20, 2006 Report Share Posted December 20, 2006 Obviously Dave, Sebastian even said so in #1, the puzzle is to do it "without knowing that" by just treating it as an equation. Quote Link to comment Share on other sites More sharing options...

Nootropic Posted January 17, 2007 Report Share Posted January 17, 2007 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! CraigD 1 Quote Link to comment Share on other sites More sharing options...

max4236 Posted January 17, 2007 Report Share Posted January 17, 2007 a^3+b^3=9 thanks :) Quote Link to comment Share on other sites More sharing options...

cwes99_03 Posted January 17, 2007 Report Share Posted January 17, 2007 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=? ? Quote Link to comment Share on other sites More sharing options...

Nootropic Posted January 17, 2007 Report Share Posted January 17, 2007 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. Quote Link to comment Share on other sites More sharing options...

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