@Richiez

On behalf of all the participants, I say a big THANK YOU to you, Maximus, etc.

The sacrifice you people have put into this competition can best be imagined.

THANK YOU ALL!!!

Hay guys, drop ur 2go usernames..or any oda i.d.

smurfy:

Oh! You mean you used trial and error! See me o! I was trying to solve an equation that contains sixth root!

Alpha will post the solution,I just tried substituting some assumed values into the equations I got and finally got the correct answer!

Thanks for the challenge so far. You are truly a great Mathematician! Send me a PM,let's get to know each other better.

jaryeh:

LMAO

Congrats. ___________0__

smurfy: @Richiez

On behalf of all the participants, I say a big THANK YOU to you, Maximus, etc.

The sacrifice you people have put into this competition can best be imagined.

THANK YOU ALL!!!

Signed: jaryeh

Nairaland physics quiz ..coming up. After ASUU strike..

denniece1:

Congrats. ___________0__

Thanks Oga mi

A dream of yesterday has finally turned out to be reality.

Thanks to our panel of judges
Olawalebabs
Fynestboi

Thanks to the Quiz masters

Doubledx
[url=https://www.nairaland.com/alpha%20maximus]AlphaMaximus[/url]


Thanks to all Participants

Thanks to all who gave us advice, support, and to all who contributed one way or the other, especially our audience

Thanks everyone! and until we gather next time for the next edition I remain


Richiez

Mathematically transforming all aforementioned details concerning the relationship of the two unknown numbers, we have that:
XY=1,000,000.....(I)
X^1/6=Y^1/2 -3.....(II)
This can be solved logarithmically, thus
Logxy=Log1,000,000
LogX + LogY=6.......(III)
Logx^1/6= Log(y^1/2-3)
1/6Logx=Log(y^1/2-3)
Logx=Log(y^1/2-3)^6
Logx=6Log(y^1/2-3).....(IV)
Substitute (IV) into (III)
6Log(y^1/2-3) + LogY=6
Divide through by 6 for simplicity
Log(y^1/2-3) +1/6 Logy=1
Log(y^1/2-3)+ Logy^1/6
Applying the Log law, log.a + log.b=log.(AB), we have:
Log(Y^1/6(y^1/2 - 3))= 1
Eradicating the logarithm, we have that:
Y^1/6(Y^1/2-3)=10^1=10
Expanding we have:
Y^2/3-3y^1/6-10=0
Applying algebraic substitution, I will transform Y^1/6 to 'a'
NB: Y^2/3=(Y^1/6 )^4
A^4-3A-10=0
Using trial and error, it is found that A=2,
Comparing the result with Y^1/6, we have:
Y^1/6=2
Y=2^6=64
Y= 64 , substituting y=64 back to eq(I) we have: 64x=1,000,000 , x=1,000,000/64=15,625 Thus x=15,625 and y=64......these are the two unknown numbers, interestingly they are simultaneously perfect squares and perfect cubes, they are known as perfect sixth powers, but this is not the final solution:
The cube root of the larger number 15,625 is 25 and the sixth root(number multiplied six times yielding the cited figure) of 64 is 2...their product is 25*2=50 the reciprocal is 1/50 and the square of this is 1/2500........oya clap for me!!

congrats JARYEH our Maths genius

Dane17: congrats JARYEH our Maths genius

Thanks!

Congrats Jaryeh!!!

I am glad you won. I look forward to meeting you in the flesh after ASUU strike.

I also congratulate the other participants. There was always gonna be one winner, but we have all learnt from this experience.

uceee: Congrats Jaryeh!!!

I am glad you won. I look forward to meeting you in the flesh after ASUU strike.

I also congratulate the other participants. There was always gonna be one winner, but we have all learnt from this experience.

Thanks boss!! You can PM me.

Maximus... Na you OOOOOOO

So sad i could nt see this to the end, i had to rush down for an urgent rehearsal. All things been equal. Jaryeh i knw you own that trophy congrat...

jaryeh:

Thanks boss!! You can PM me.

Sent!

Let me decorate our 2013 Nairaland Mathematics guru...

[NISD) Friday match 29th NOVEMBER 2013

FUTA VS UNIJOS


UNIBADAN VS DELSU


(NISD) SATURDAY MATCH 30TH

NDU VS UNIBEN


OSUSTECH VS UNN

uceee: Sent!

seen!

Fynestboi: So sad i could nt see this to the end, i had to rush down for an urgent rehearsal. All things been equal. Jaryeh i knw you own that trophy congrat...

Oga mi! Thank you sir!

jaryeh:

Oga mi! Thank you sir!

handshake.

Nice workings so far and great job well done by the Mods and cordinators along with the quiz masters.

Great showings from all participants and a warm congrats to Jaryeh a homie. #TeamOAU. Nice one bro.

Mr Calculus: WONDERFUL

@Boss, please contact me on trustyondstreets@gmial.com // I got your NL message but it couldn't reply.

Alpha Maximus: Mathematically transforming all aforementioned details concerning the relationship of the two unknown numbers, we have that:
XY=1,000,000.....(I)
X^1/6=Y^1/2 -3.....(II)
This can be solved logarithmically, thus
Logxy=Log1,000,000
LogX + LogY=6.......(III)
Logx^1/6= Log(y^1/2-3)
1/6Logx=Log(y^1/2-3)
Logx=Log(y^1/2-3)^6
Logx=6Log(y^1/2-3).....(IV)
Substitute (IV) into (III)
6Log(y^1/2-3) + LogY=6
Divide through by 6 for simplicity
Log(y^1/2-3) +1/6 Logy=1
Log(y^1/2-3)+ Logy^1/6
Applying the Log law, log.a + log.b=log.(AB), we have:
Log(Y^1/6(y^1/2 - 3))= 1
Eradicating the logarithm, we have that:
Y^1/6(Y^1/2-3)=10^1=10
Expanding we have:
Y^1/3-3y^1/6-10=0
Applying algebraic substitution, I will transform Y^1/6 to 'a'
NB: Y^1/3=Y^1/6 * Y^1/6
A^2-3A-10=0
(A-5)(a+2)=0
Thus a=5 and a=-2
Compare the results with Y^1/6
5=Y^1/6
Y=5^6=15,625
Also, Y^1/6=-2
Y=-2^6=64
Y= 64 and 15,625......these are the two unknown numbers, interestingly they are simultaneously perfect squares and perfect cubes, they are known as perfect sixth powers, but this is not the final solution:
The cube root of the larger number 15,625 is 25 and the sixth root(number multiplied six times yielding the cited figure) of 64 is 2...their product is 25*2=50 the reciprocal is 1/50 and the square of this is 1/2500........oya clap for me!!

Y^1/6 (Y^1/2 - 3) = Y^2/3 - 3 Y^1/6

There's a mistake in your expansion

Appliedmaths.:


@Boss, please contact me on trustyondstreets@gmial.com // I got your NL message but it couldn't reply.

ALREADY DON SIR

papindinho: Nice workings so far and great job well done by the Mods and cordinators along with the quiz masters.

Great showings from all participants and a warm congrats to Jaryeh a homie. #TeamOAU. Nice one bro.

Thanks bro!

Congrat to all the finalists and most especially the winner-jaryeh. I am proud of you all! Please richiez, don't leave me out on this next time (season 2).

jaryeh:

Thanks boss!! You can PM me.

my boy Jaryeh!

Am very proud of you!

The sky is your starting point!

@Jaryeh

Oga mi sir,


Congrats!

Congrats jayreh
pls ah shouldn't be left out in d next reason o

Tolexander: my boy Jaryeh!

Am very proud of you!

The sky is your starting point!

Oga mi,thanks sir!

factorial1: @Jaryeh

Oga mi sir,


Congrats!

Big boss! Thanks sir!