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

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
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

guy 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

beatsbyj2g: guy 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

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?

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
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

I solved this question sometime ago on Nairaland Maths Clinic thread. Check there or check my previous posts.

jaryeh:

I solved this question sometime ago on Nairaland Maths Clinic thread. Check there or check my previous posts.

kindly forward a link to this thread or the post...thanks

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
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

gbengarock:

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?

have tried using log it didn't work wat else na

thankyouJesus: Use graphical method of solving equation

bros i go jst go learn dat one o...is dere no oda way?

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

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
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

m

Trial and error method.

gbengarock:

bros i go jst go learn dat one o...is dere no oda way?

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.

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.

actYourDreams: 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.

fnks boss

gbengarock:

kindly forward a link to this thread or the post...thanks

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.

[quote author=gbengarock post=26750408] what are happy end theorem.....

actYourDreams:
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.

use linear approximation.....

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.