Am still expecting solution to 3^x + 4^x = 5^x

@ d citizen
i have the solution to ur questiön 3^x+4^x=5^x. Buh is a graphical solution which indicate dat x=2. I dnt knw hw to paste it ön nairaland.

@ d citizen.
From y=x^x^x. dy/dx=x^x^x[x^x(lnx+1)lnx+x^x/x]. Now y=x^(x^x^x).
take log of both sidés.
lny=lnx^(x^x^x).
lny=x^x^xlnx.
d/dx(lny)=d/dx(x^x^xlnx).
dy/dx(1/y)=udv/dx+vdu/dx. Take u=x^x^x du/dx=x^x^x[x^x(lnx+1)lnx+x^x/x]. And v =lnx dv/dx=1/x.
dy/dx=y[x^x^x/x+lnx(x^x^x[x^x(lnx+1)lnx+x^x/x].
now recall that y=x^x^x^x. Therefore dy/dx=x^x^x^x[x^x^x/x+lnx(x^x^x[x^x(lnx+1)lnx+x^x/x].
similarly for y=x^(x^x^x^x) Take log of both sides.
lny=x^x^x^xlnx
d/dx(lny)=d/dx(x^x^x^xlnx). Take u=x^x^x^x du/dx=x^x^x^x[x^x^x/x+lnx(x^x^x[x^x(lnx+1)lnx+x^x/x]. And take v=lnx dv/dx=1/x . Using product rule. We av
(1/y)dy/dx=x^x^x^x/x+lnx(x^x^x^x[x^x^x/x+lnx(x^x^x[x^x(lnx+1)lnx+x^x/x].
dy/dx=y[x^x^x^x/x+lnx(x^x^x^x[x^x^x/x+lnx(x^x^x[x^x(lnx+1)lnx+x^x/x]]. But y=x^x^x^x^x. Therefore dy/dx=x^x^x^x^x([x^x^x^x/x+lnx(x^x^x^x[x^x^x/x+lnx(x^x^x[x^x(lnx+1)lnx+x^x/x]]). For easier understanding let y=X^x^x^x^x.
u=x^x^x^x.
v=x^x^x.
w=x^x. Hence
dy/dx=yu/x+uvy[(w(lnx+1)lnx]+w/x+w/x. All is well.

Rhydex the solution to 3^x + 4^x = 5^x can be done analytically and graphically. What i need from is the analytical solution since u claim to be a five star math general

Mathematics is not as daunting as it seems,it's all about following simple rules.Repeated use of these rules builds your understanding and confidence .

Angwe Hemen: Mathematics is not as daunting as it seems,it's all about following simple rules.Repeated use of these rules builds your understanding and confidence .

#applauds# Welcome!

Please guru in the house, I'm finding abstract algebra very difficult. Can someone give me the rudiment toward understanding it?

hmm

Good evening to you all.I actually thought i was obeying the rules ,parallel to my thought..i was unable to post for the past 48hrs...master richiez..please what went wrong.?

Richiez: sure! anytime u'r ready

Please help me solve dis question.... 5/12 of a number is subtracted from 3/4 of the number. The positive difference is 7 less than 5/6 of the number. Find the number

Pls, help me with the solution of this question with its explanation:

The difference between the digits of a two-digit number is 1. The number itself is 1 more than 5 times the sum of its digits. If the units digit is greater than the tens digit, find the number.

5/12 of a number is subtracted from 3/4 of the number. The positive difference is 7 less than 5/6 of the number. Find the number. plz help me

Odehfamily: 5/12 of a number is subtracted from 3/4 of the number. The positive difference is 7 less than 5/6 of the number. Find the number. plz help me

The number is -18 1/2

busuyem: Pls, help me with the solution of this question with its explanation:

The difference between the digits of a two-digit number is 1. The number itself is 1 more than 5 times the sum of its digits. If the units digit is greater than the tens digit, find the number.

Still expecting the solution pls.

busuyem:

The number is -18 1/2

Plz show me the working.

Integrate me.

Fetus: X+y =5
x^x + y^y=31.......find x and y....

...it's obvious that the answer is (3,2) or (3,2) as (x,y)...but if it's of no simple root...use newton raphson method...that x'=x-f(x)/f'(x)...i also heard that the great newton was unable to solve the question

benbuks: Good evening to you all.I actually thought i was obeying the rules ,parallel to my thought..i was unable to post for the past 48hrs...master richiez..please what went wrong.?

It was the antispam bot that got you banned...although it's very efficient, it wrongly bans members sometimes...I had to intervene ...sorry for the inconvenience anyway

d citizen:
Am still expecting solution to 3^x + 4^x = 5^x

i have two solutions but i'll start wit d easier one. Let # denote "congruency" then since 3^x+4^x=5^x, 4^x#5^x(mod3) or by Fermat's Little theorem (4^x)^2#(5^x)^2#1(mod3), i.e 4^2x#5^2x#1(mod3) or 16^x#1(mod3) and 25^x#1(mod3) so by FLT, x must divide @(3)=3-1=2 Euler's phi function, i.e x|2 hence either x=1 or 2 but since 1 does not satisfy the eqn, x=2,

d citizen:
Differentiate y= x^x^x^x^x to infinte with respect to x . I pose dis question to the math general

...y=x^x^x^x^x to infinity...which gives y={x^(x^4)}^infinity....but x^4*infinity=infinity...then,y=x^infinity...dy/dx=infinity.x^infinity-1...dy/dx=infinity.x^infinity...since infinity-1=infinity...then dy/dx=infinity...

Use fermat last theorem (FLT)..TO solve It..

busuyem: Help solve dis o-the numerator of a fraction is 5 less than its denominator.if 6 is added to the numerator and 4 to the denominator,the fraction is doubled.what is the fraction?Pls, help me out with this:

...let the denominator be x...since the numerator is 5 less than the denominator...then the original fraction is (x-5)/x...but the question continues that 6 is added to the numerator and 4 to the denominator...then (x-5+6)/(x+4)...which is equal to 2 times the original equation{(x-5)/x}...then (x+1)/(x+4)=2(x-5)/x...multiply through by x(x+4){which is the same as cross multiplication} to give x(x+1)=2(x-5)(x+4)...x^2+x=2x^2-2x-40...x^2-3x-40....x=8 or -5...since the original equation is (x-5)/x...substitute 8 which gives 3/8 OR substitute -5 which gives 2

Odehfamily:
Please help me solve dis question.... 5/12 of a number is subtracted from 3/4 of the number. The positive difference is 7 less than 5/6 of the number. Find the number

Okay here it is,
Let the Number be X.
(3/4)X - (5/12)X = (5/6)X - 7
Collect like terms
(3/4)X - (5/12)X - (5/6)X = -7
-(6/12)X = -7
-(1/2)X = -7
X= 14

Calculusf(x):
...y=x^x^x^x^x to infinity...which gives y={x^(x^4)}^infinity....but x^4*infinity=infinity...then,y=x^infinity...dy/dx=infinity.x^infinity-1...dy/dx=infinity.x^infinity...since infinity-1=infinity...then dy/dx=infinity...

,.i think question has already bin solved.

Richiez:
It was the antispam bot that got you banned...although it's very efficient, it wrongly bans members sometimes...I had to intervene ...sorry for the inconvenience anyway

...its ok..thanks man....lets kip rollin.

I can see.my generals in d house....i dey project salute oo.

Odehfamily: 5/12 of a number is subtracted from 3/4 of the number. The positive difference is 7 less than 5/6 of the number. Find the number. plz help me

...let the number be x...(3x/4)-(5x/12)=(5x/6)-7...(9x-5x)/12=(5x-42)/6...4x=10x-84...x=14

For the benbuks question....y=∫√1+√x.dx.....let √x be a²...then x=a^4...dx=4a^3da...substitute to give ∫4a^3√1+a².da....let u be 1+a²...du/2a=da...then ∫4a²√u...4∫a²√u but u-1=a²...then,4∫(u-1)√u...4∫(u^3/2)-(u^1/2)...4{[2u^(5/2)]/5-[2u^(3/2)]/3}...8/5.u^(5/2)-8/3.u^(3/2)...but u=a²+1 and a^4=x...substitute those

Calculusf(x):
For the benbuks question...£ is my integral sign.y=∫√1+√x.dx.....let √x be a²...then x=a^4...dx=4a^3da...substitute to give ∫4a^3√1+a².da....let u be 1+a²...du/2a=da...then ∫4a²√u...4∫a²√u but u-1=a²...then,4∫(u-1)√u...4∫(u^3/2)-(u^1/2)...4{[2u^(5/2)]/5-[2u^(3/2)]/3}...8/5.u^(5/2)-8/3.u^(3/2)

....u try bro..bt it seems. U didnt get d question....integrate sqrt[(1 +sqrt(x)]/x..dats d question...re-try......nice 1 dia "calculus(fx)".

Calculusf(x):
For the benbuks question...£ is my integral sign.y=∫√1+√x.dx.....let √x be a²...then x=a^4...dx=4a^3da...substitute to give ∫4a^3√1+a².da....let u be 1+a²...du/2a=da...then ∫4a²√u...4∫a²√u but u-1=a²...then,4∫(u-1)√u...4∫(u^3/2)-(u^1/2)...4{[2u^(5/2)]/5-[2u^(3/2)]/3}...8/5.u^(5/2)-8/3.u^(3/2)

guy u said u got admission to read mechanical engr......but how come u know all this.....did u do any A-LEVEL program?

emmyeuler1: guy u said u got admission to read mechanical engr......but how come u know all this.....did u do any A-LEVEL program?

...i'm still your boy master...i haven't done any A-level work ooo.....i pray God helps

benbuks: ....u try bro..bt it seems. U didnt get d question....integrate sqrt[(1 +sqrt(x)]/x..dats d question...re-try......nice 1 dia "calculus(fx)".

...y=∫√(1+√x)/x.dx...let √x=a²...x=a^4...dx=4a^3.da...then,∫√(1+a²)/a^4.4a^3da...∫√(1+a²)/a..da...if u²=1+a²...udu=ada...da=udu/a...then ∫u/a*udu/a...∫u²/a².du but u²=1+a²...a²=u²-1...then ∫u²/(u²-1).du...do the division to get...∫1+1/(u²-1).du...from here...let's solve 1/u²-1...resolve to partial fraction,1/u²-1=(A/u-1)+(B/u+1)....1=A(u+1)+B(u-1)...if u=1...A=1/2...and B=-1/2....which gives,1/2(u-1)-1/2(u+1)...then,∫1+1/2(u-1)-1/2(u+1).du...then,{u+1/2ln(u-1)-1/2ln(u+1}...{u+1/2ln(u-1)/(u+1)}+c...don't forget that u²=1+a²...u=√1+a²...don't forget that a²=√x...u=√1+√x...substitute that to the answer to give...{√(1+√x)+(1/2)ln(√1+√x-1)/(√1+√x+1)+c...{√(1+√x)+(1/2)ln(x^1/4)/(√2+√x)+c...i guess.....waiting for better solutions from my generals