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Pls Maths Gurus In The House, Help / Help A Brother Solve Some Shit / Maths Gurus Solve This Using BODMAS: 6 Divide By 2(3+2) (2) (3) (4)
Help A Brother Solve A Problem, Maths Gurus! by gbengarock: 12:39pm On Sep 30, 2014 |
Tho' simple it seems but i've been on this since i was in secondary school. It's a problem we created for ourselves by removing the square(power of 2) and substituting with 'x' in the pythagoras triple As we all know that 3^2+4^2=5^2 note: ^2 denotes 'raised to power of 2' Question: prove that x=2 in 3^x+4^x=5^x thank you |
Re: Help A Brother Solve A Problem, Maths Gurus! by beatsbyj2g(m): 1:24pm On Sep 30, 2014 |
gbengarock: Tho' simple it seems but i've been on this since i was in secondary school. It's a problem we created for ourselves by removing the square(power of 2) and substituting with 'x' in the pythagoras tripleguy stop messing with mathematics normally 3^2+4^2 is not = 5^2 which is 25 But any if u have d answer show me how u solved it |
Re: Help A Brother Solve A Problem, Maths Gurus! by gbengarock: 3:38pm On Sep 30, 2014 |
beatsbyj2g: guy stop messing with mathematics normally 3^2+4^2 is not = 5^2 which is 25 i seem to be learnin something new from you here. Wat do u mean by 3^2+4^2 is not = 5^2 if 3^2=9, 4^2=16 and 5^2=25. is 9+16 not = 25? |
Re: Help A Brother Solve A Problem, Maths Gurus! by jaryeh(m): 3:46pm On Sep 30, 2014 |
gbengarock: Tho' simple it seems but i've been on this since i was in secondary school. It's a problem we created for ourselves by removing the square(power of 2) and substituting with 'x' in the pythagoras triple I solved this question sometime ago on Nairaland Maths Clinic thread. Check there or check my previous posts. |
Re: Help A Brother Solve A Problem, Maths Gurus! by gbengarock: 4:07pm On Sep 30, 2014 |
jaryeh: kindly forward a link to this thread or the post...thanks |
Re: Help A Brother Solve A Problem, Maths Gurus! by thankyouJesus(m): 5:19pm On Sep 30, 2014 |
Use graphical method of solving equation gbengarock: Tho' simple it seems but i've been on this since i was in secondary school. It's a problem we created for ourselves by removing the square(power of 2) and substituting with 'x' in the pythagoras triple |
Re: Help A Brother Solve A Problem, Maths Gurus! by beatsbyj2g(m): 5:25pm On Sep 30, 2014 |
gbengarock:have tried using log it didn't work wat else na |
Re: Help A Brother Solve A Problem, Maths Gurus! by gbengarock: 5:51pm On Sep 30, 2014 |
thankyouJesus: Use graphical method of solving equation bros i go jst go learn dat one o...is dere no oda way? |
Re: Help A Brother Solve A Problem, Maths Gurus! by gbengarock: 5:53pm On Sep 30, 2014 |
beatsbyj2g: have tried using log it didn't work wat else na i've also tried all of that...i jst wish someone wud help |
Re: Help A Brother Solve A Problem, Maths Gurus! by actYourDreams: 6:16pm On Sep 30, 2014 |
Your problem was not well-defined. Math is a concise language so when posing a math problem you need to be precise. By saying "prove that x = 2", I will assume that your solution has to be a natural number. Hence, my following approach is based on the assumption that x must be a natural number. Observe that 4^x = 2^(2x) and 5^x = (3+2)^x. Therefore your problem can be written as 2^(2x)+ 3^x=(3+2)^x. Then apply binomial expansion to (3+2)^x, i.e. (3+2)^x = 3^x + 2x3^(x-1) + (x(x-1))/2 * 3^(x-2)2^2 + ... + x*3*2^(x-1) + 2^x. Therefore the question becomes 2^(2x) = 2x3^(x-1) + (x(x-1))/2 * 3^(x-2)2^2 + ... + x*3*2^(x-1) + 2^x, since the 3^x on both sides cancel out. Then it easily follows that for the previous equality to hold x must be 2 or use mathematical induction to show that. If NL had Latex capability I might have clarified further. However, Ihope my approach was clear enough. gbengarock: Tho' simple it seems but i've been on this since i was in secondary school. It's a problem we created for ourselves by removing the square(power of 2) and substituting with 'x' in the pythagoras triple |
Re: Help A Brother Solve A Problem, Maths Gurus! by actYourDreams: 6:18pm On Sep 30, 2014 |
m |
Re: Help A Brother Solve A Problem, Maths Gurus! by thankyouJesus(m): 7:19pm On Sep 30, 2014 |
Trial and error method. gbengarock: |
Re: Help A Brother Solve A Problem, Maths Gurus! by thankyouJesus(m): 7:23pm On Sep 30, 2014 |
I raise my cap for the boss actYourDreams: Your problem was not well-defined. Math is a concise language so when posing a math problem you need to be precise. |
Re: Help A Brother Solve A Problem, Maths Gurus! by gbengarock: 3:18pm On Oct 01, 2014 |
actYourDreams: Your problem was not well-defined. Math is a concise language so when posing a math problem you need to be precise.fnks boss |
Re: Help A Brother Solve A Problem, Maths Gurus! by jaryeh(m): 3:46pm On Oct 01, 2014 |
gbengarock: It seems the post was wiped off by NL tsunami. Well, this is the approach: 3^x = (1+2)^x, and by binomial expansion 3^x = xC0.1^x.2^0 +xC1.1^(x-1).2^1 + xc2.1^(x-2).2^2 + .....You can stop here since, by mere inspection, x is very small. Simplify that, then express others in that form, simply also, then solve - it should result in a quadratic equation. |
Re: Help A Brother Solve A Problem, Maths Gurus! by donfourier(m): 11:37am On Jan 11, 2015 |
[quote author=gbengarock post=26750408] what are happy end theorem..... |
Re: Help A Brother Solve A Problem, Maths Gurus! by donfourier(m): 11:44am On Jan 11, 2015 |
actYourDreams:use linear approximation..... |
Re: Help A Brother Solve A Problem, Maths Gurus! by bolkay47(m): 2:24pm On Jan 11, 2015 |
I think we can use newton ralphson method f(x)=3^x+4^x-5^x f“(x)=3^xln3+4^xln4-5^xln5 use x0=0 as starting point.... ALSO 3^x+4^x=5^x divide through by 4^x (3/4)^x+1=(5/4)^x (0.75)^x+1=(1.25)^x (1-0.25)^x+1=(1+0.25)^x Then use expansion of a sum.... You will arrive at 1.96875-0.25x+0.03725x^2=0.96875+0.25x+0003725x^2 Collect the like terms 1=0.5x X=2. |
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