2006-08-24 00:16:12 · 7 answers · asked by gara_to 1 in Education & Reference ➔ Other - Education
If you play with the numbers you get x=4; y=1
Its called the guess test and revise,, it works best here
since you know that anything raised to the 1power give you back that same number you know sqr root X has to equal 2
So X must be 4 √4^1=2
so 4^3(1)= 64
You can solve this by taking the log of both sides but that just makes it really difficult
y log√x= log 2
y= log2/ log√x
substitue that into
the other eqn and solve for X
x^3( log2/log√x=0
once you find X
substitue X back into the original eqn and solve for Y
Then you will have 2 eqn for Y and will be able to set them equal to each other it get a specific value..Then do the same for X
2006-08-24 00:22:58 · answer #1 · answered by ۞ JønaŦhan ۞ 7 · 3⤊ 0⤋
it is 2^6 =64
2006-08-24 00:26:14 · answer #2 · answered by Kumar 5 · 0⤊ 0⤋
yea, 64
as u know (âx)^2 = x so
x^3y = (âx)^2^3y then
= (âx)^6y = (âx)^y^6 = 2^6 = 64
2006-08-24 00:25:37 · answer #3 · answered by Shervin 2 · 0⤊ 0⤋
Um......64.
2006-08-24 00:19:48 · answer #4 · answered by JeffE 6 · 0⤊ 0⤋
Hmmmm... not sure what to do with the bit left over......
2006-08-24 00:23:22 · answer #5 · answered by maggie rose 4 · 0⤊ 0⤋
x^(6ln2/lnx)
2006-08-24 00:17:58 · answer #6 · answered by Elim 5 · 0⤊ 0⤋
penis
2006-08-24 00:18:11 · answer #7 · answered by The Samster 2 · 0⤊ 0⤋