₦airaland Forum

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

There's a mistake in your expansion

Yep, I've seen it!Thanks for noticing!! It has been corrected now and is as flawless as can be!! Its amazing how I still arrived at the right answer with that 'small' mistake...

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!!

Solution loading

smurfy: Maximus! Maximus!! Maximus!!!

What did I say about you setting this humongous type of questions?

... It had to be done my good man, it had to be done!!!

jaryeh: 1/2500

...CORRECT!!!!

Why the delay, the question is not that hard!

benbuks: alpha the killer.

Richiez: Quiz master alphamaximus pls check your mail

Seen

Loading question

Question 4
Find the values of x and y that satisfy the simultaneous equations:
7x-4y=23
4x-3y=11
A.2,3 B. 5,3 C. 3,3 D.5,2

Question 8

The limit of the function (Sinx)/x as 'x' tends to zero is?

A. 0 B. X C. 1 D. Ask Google

Question 11

Find the area enclosed by the curve, Y=25- x² and the straight line, Y=x+13

A. 57.2 B. 52.7 C. 58.2 D. 25.2

Loading question...

smurfy: He never said first to post.

... He didn't have to say it. From the logical point of view, who do you think deserves it? Won't it be the person who did it first?

Question 2
Solve the following for x: 2Log(sub)a(/sub)x-Log(sub)a(/sub)x-1=Log(sub)a(/sub)x-2
A. X=3/2 B. X=1/3 C. X=2 D. X=2/3

@ Richiez, pls check your mail

Hmmmm, the foundations of mathematics shall be shaken today! The mathematical bones of our participants shall be rigorously worked to the marrow Goodluck to you all, I sincerely hope at least one of my questions can be solved

Solution:
1/2(x+2)+1/3(x-3)=20
Multiply thru by 6
3x+6+2x-6=120
5x=120
X=120/5=24