help my 10th grade sibling
08-04-2009, 09:52 PM
#1
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help my 10th grade sibling i went all the way in to calculus in high school and i cant fucken remember this algebra problem 
x=32^(1/5)
x^(5/1)= 32^[(1/5)*(5/1)]
x^5=32
i know x=2 cuz the back of the book said it and i verified with:
2*2*2*2*2
..4*4*2
....16*2
......32
but how would you do it with out knowing already that x=2
how would you solve it step by step...
x=32^(1/5)
x^(5/1)= 32^[(1/5)*(5/1)]
x^5=32
i know x=2 cuz the back of the book said it and i verified with:
2*2*2*2*2
..4*4*2
....16*2
......32
but how would you do it with out knowing already that x=2
how would you solve it step by step...
08-04-2009, 10:19 PM
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what more do you need to know looks like you have it down, except for maybe the fact that not too many people can do fifth roots off the top of their head. This isnt too much of a algebra problem really, more just understanding powers and roots.
i.e. X= 32^(1/5) = 5th root (32)
the reason it usually is shown as 32^(1/5) is that unless you have a program such as mathtype keyboards dont have a sqaure root sign or in this case a fifth root sign. Here is a janky looking version of the root sign, in which you would place a five in the "v" portion to denote the fifth root of something.
( \/``` )
Then the all mighty calculator come into play.
i.e. X= 32^(1/5) = 5th root (32)
the reason it usually is shown as 32^(1/5) is that unless you have a program such as mathtype keyboards dont have a sqaure root sign or in this case a fifth root sign. Here is a janky looking version of the root sign, in which you would place a five in the "v" portion to denote the fifth root of something.
( \/``` )
Then the all mighty calculator come into play.
08-04-2009, 10:22 PM
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i went all the way in to calculus in high school and i cant fucken remember this algebra problem
x=32^(1/5)
x^(5/1)= 32^[(1/5)*(5/1)]
x^5=32
i know x=2 cuz the back of the book said it and i verified with:
2*2*2*2*2
..4*4*2
....16*2
......32
but how would you do it with out knowing already that x=2
how would you solve it step by step...
x=32^(1/5)
x^(5/1)= 32^[(1/5)*(5/1)]
x^5=32
i know x=2 cuz the back of the book said it and i verified with:
2*2*2*2*2
..4*4*2
....16*2
......32
but how would you do it with out knowing already that x=2
how would you solve it step by step...
Mrs. 2c5s says:
It means that you are taking the fifth root of 32; meaning what number is the base if the exponent is 5 and the result is 32. It's a square root problem.
Hope that makes sense.
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08-04-2009, 10:25 PM
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what more do you need to know looks like you have it down, except for maybe the fact that not too many people can do fifth roots off the top of their head. This isnt too much of a algebra problem really, more just understanding powers and roots.
i.e. X= 32^(1/5) = 5th root (32)
the reason it usually is shown as 32^(1/5) is that unless you have a program such as mathtype keyboards dont have a sqaure root sign or in this case a fifth root sign. Here is a janky looking version of the root sign, in which you would place a five in the "v" portion to denote the fifth root of something.
( \/``` )
Then the all mighty calculator come into play.
i.e. X= 32^(1/5) = 5th root (32)
the reason it usually is shown as 32^(1/5) is that unless you have a program such as mathtype keyboards dont have a sqaure root sign or in this case a fifth root sign. Here is a janky looking version of the root sign, in which you would place a five in the "v" portion to denote the fifth root of something.
( \/``` )
Then the all mighty calculator come into play.
thank Mrs 2c5s
its just that i didnt want her to use the calculator, i wanted her to do it free hand first
08-04-2009, 10:51 PM
#10
x^5=32
what times itself 5 times = 32?
2^5 = 32 (2*2*2*2*2)
x^4= 16
what number times itself 4 times = 16?
2^4 = 16 (2*2*2*2)
when its simplified... SQUARE ROOT... x^2 = (a) ~a= whatever variable
what number times itself (2 times) = a
if a = 4 , x = 2 (square root of 4)
if a = 400 x = 20 (square root of 400)
CUBED ROOT x^3 = (a)
if a=3, x=1... Right? No... 1^3= 1*1*1 = 1 ... 1 is not = 3
So **if were staying with whole numbers**, a cannot = 3.
a=27, x=3 ... 3^3 = 27 ... Yes. (3*3*3)
Its just difficult to start thinking of things in powers over 3, because we just dont use it much. All its asking is what number, mulitplied _ times = your variable.
x= answer
y= root
a= number
x^y=a with any 2, you should be able to find the 3rd. I think in algerbra you stick with whole numbers, so you wont have (example) x^3 = 3 as it would be a decimal answer... actually i think it would be a limit (calc 1 and 2).
*=> (means is = to)
x^y=a => x^(1/x)=y
3^2=6 => 6^(1/3)=2
So you could get...
3^y=6 => 6^(1/3)= y ; y=2
or
x^2=6 => 6^(1/x)=2 ; x=3
or, the most common...
3^2=a => a^(1/3)=2 ; a=6
Therefore you know that anytime you see
a^(1/x)=y ...you should be looking to convert it to:
x^y=a ...because the calculations ae easier to comprehend.
what times itself 5 times = 32?
2^5 = 32 (2*2*2*2*2)
x^4= 16
what number times itself 4 times = 16?
2^4 = 16 (2*2*2*2)
when its simplified... SQUARE ROOT... x^2 = (a) ~a= whatever variable
what number times itself (2 times) = a
if a = 4 , x = 2 (square root of 4)
if a = 400 x = 20 (square root of 400)
CUBED ROOT x^3 = (a)
if a=3, x=1... Right? No... 1^3= 1*1*1 = 1 ... 1 is not = 3
So **if were staying with whole numbers**, a cannot = 3.
a=27, x=3 ... 3^3 = 27 ... Yes. (3*3*3)
Its just difficult to start thinking of things in powers over 3, because we just dont use it much. All its asking is what number, mulitplied _ times = your variable.
x= answer
y= root
a= number
x^y=a with any 2, you should be able to find the 3rd. I think in algerbra you stick with whole numbers, so you wont have (example) x^3 = 3 as it would be a decimal answer... actually i think it would be a limit (calc 1 and 2).
*=> (means is = to)
x^y=a => x^(1/x)=y
3^2=6 => 6^(1/3)=2
So you could get...
3^y=6 => 6^(1/3)= y ; y=2
or
x^2=6 => 6^(1/x)=2 ; x=3
or, the most common...
3^2=a => a^(1/3)=2 ; a=6
Therefore you know that anytime you see
a^(1/x)=y ...you should be looking to convert it to:
x^y=a ...because the calculations ae easier to comprehend.
Last edited by xxrillixx; 08-04-2009 at 11:19 PM.
08-05-2009, 12:03 AM
#14
And you were right here... just backwards.
x^5=32
32
option 1: 2*(16) (2 numbers)
2*(4*4) (3 numbers)
2*(2*2*2*2) (5 numbers) - Thats what we want!
32
option 2: (4)*(8)
(2*2)*(2*2*2)
**Note both option 1 and option 2 turned out the same.
What does this mean... It means you cannot do 32 in any other root. Why not? Because you dont have any number multiplied only by itself. (if were still using whole numbers**)
x^2= 32
32
option 1: 4*8 - no
32
option 2: 2*16 - no
------------------
x^3=32 (looking for a number multiplied by itself 3 times that equals 32)
32
option 1: 4*8
option 1a: (2*2)*8 - no
option 1b: 4*(4*2) - no
2: 2*16
2a: 2*(2*8) - no
2b: 2*(4*4) - no
There can be MANY MANY 'options'...
x^4=32 (looking for a number multiplied by itself 4 times that equals 32)
32
1: 2*2*8
(2*2)*(2*4) - not possible
But take a number like 64
x^3=64
64
4*16
4*4*4 - x=4 But wait.... Why can i reduce this further?
(2*2)*(2*2)*(2*2) - What does this mean?
it means that 2*2*2*2*2*2 = 64... or 2^6=64
try 256
16*16 - 16^2=256
(4*4)(4*4) - 4^4=256
(2*2*2*2)(2*2*2*2) - 2^8=256
What else does this tell us?
16^2=256
4^4=256
2^8=256
Well we know that: 16^2 = 4^4 = 2^8 ... right?
so x^y = square root(x)^(y)*2 = 4th root(x)^y(4)
so using x=16 , y=2
16^2 = 256
square root (16)^(2)*2 = 4^4 = 256
4th root (16)^(2)*4 = 2^8 = 256
Lets apply that to the number earlier 32 2^5 ; x^y = square root(x)^(y)*2 = quard root(x)^y(4) ... SR= square root
x=2, y=5
2^5 = 32
square root(2)^(5)*2 - not possible
BECAUSE the SR(2) is a decimal ... we are only dealing with whole numbers. So we proved again that 32 cannot have any other possible # of roots. (IE x^3 cannot = 32 , x^2 cannot =32 , because it creates decimals.)
x^5=32
32
option 1: 2*(16) (2 numbers)
2*(4*4) (3 numbers)
2*(2*2*2*2) (5 numbers) - Thats what we want!
32
option 2: (4)*(8)
(2*2)*(2*2*2)
**Note both option 1 and option 2 turned out the same.
What does this mean... It means you cannot do 32 in any other root. Why not? Because you dont have any number multiplied only by itself. (if were still using whole numbers**)
x^2= 32
32
option 1: 4*8 - no
32
option 2: 2*16 - no
------------------
x^3=32 (looking for a number multiplied by itself 3 times that equals 32)
32
option 1: 4*8
option 1a: (2*2)*8 - no
option 1b: 4*(4*2) - no
2: 2*16
2a: 2*(2*8) - no
2b: 2*(4*4) - no
There can be MANY MANY 'options'...
x^4=32 (looking for a number multiplied by itself 4 times that equals 32)
32
1: 2*2*8
(2*2)*(2*4) - not possible
But take a number like 64
x^3=64
64
4*16
4*4*4 - x=4 But wait.... Why can i reduce this further?
(2*2)*(2*2)*(2*2) - What does this mean?
it means that 2*2*2*2*2*2 = 64... or 2^6=64
try 256
16*16 - 16^2=256
(4*4)(4*4) - 4^4=256
(2*2*2*2)(2*2*2*2) - 2^8=256
What else does this tell us?
16^2=256
4^4=256
2^8=256
Well we know that: 16^2 = 4^4 = 2^8 ... right?
so x^y = square root(x)^(y)*2 = 4th root(x)^y(4)
so using x=16 , y=2
16^2 = 256
square root (16)^(2)*2 = 4^4 = 256
4th root (16)^(2)*4 = 2^8 = 256
Lets apply that to the number earlier 32 2^5 ; x^y = square root(x)^(y)*2 = quard root(x)^y(4) ... SR= square root
x=2, y=5
2^5 = 32
square root(2)^(5)*2 - not possible
BECAUSE the SR(2) is a decimal ... we are only dealing with whole numbers. So we proved again that 32 cannot have any other possible # of roots. (IE x^3 cannot = 32 , x^2 cannot =32 , because it creates decimals.)
