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Thread: Need help with math problem
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01-18-2005, 06:03 PM #1
In this formula b = 6
a = (2 x 20^4) / b^6
So the answer is 6.8587
If I already knew that a = 6.8587, and b was the unknown, how would I solve that problem so that the answer to b didn't come out in exponential form. I mean, so that the answer would come out as "6" instead of something to the something power?
If it did come out as something to the something power, how would I convert that to a regular number?
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01-18-2005, 06:22 PM #2
Holy cow...your asking me to remember my pre calculus days. When your trying to solve to that power things get tuff. Well, I will tell you right now Im not going to do it!
Im glad thats over."If anybody can draw on the power, where do we put the meter?" - JP Morgan before pulling Tesla funding
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01-18-2005, 07:31 PM #3
Mid-
You are going to sprain your brain if you keep this up.
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01-18-2005, 07:38 PM #4
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midhvac,
Do the basic algebra and up end up with:
b=(2x20^4/a)all raised to the 1/6
You need a scientific calculator to get what b is. That is, 46656.07 ^1/6
plug in all the numbers and the answer is 6.
Is this what you were asking?
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01-18-2005, 09:53 PM #5Perhaps some number theory may be appropriate here.Originally posted by midhvac
In this formula b = 6
a = (2 x 20^4) / b^6
So the answer is 6.8587
'a' in this example is equal to exactly 320,000 / 46,656 or approximately 6.85871056241426611796982167352538 (using Microsoft calculator
)
'a' is considered a 'rational' number because it can be wriiten as a fraction, i.e., x / y where x and y are integers.
Ok, lets assume 'a' = 7, what is 'b'?Originally posted by midhvac
If I already knew that a = 6.8587, and b was the unknown, how would I solve that problem so that the answer to b didn't come out in exponential form. I mean, so that the answer would come out as "6" instead of something to the something power?
b = (320,000 / 7)^(1/6)
where 'b' is approximately 5.979643919487405903197485402773 (again using Microsoft calculator)
In this case, 'b' is an irrational number. There is no fraction, x / y, that can define 'b' where x and y are integers. Think about it! No integer numbers whatsoever. Raising this expression to the 1/6th power causes this to occur.
Mmmmm... so does this help?
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01-18-2005, 10:17 PM #6
You guys rock! Thanks! I've got a scientific calculator and the 1/6 power works.
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01-18-2005, 10:19 PM #7Nah, I put an elastic support around my head, just in case.Originally posted by condenseddave
Mid-
You are going to sprain your brain if you keep this up.
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01-18-2005, 10:46 PM #8
Hey! Wait a minute! Why are we using 1/6? Is it because the answer to b is 6? So I have to know the answer before I can work the problem?
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01-18-2005, 10:50 PM #9Let's put that elastic support to work...Originally posted by midhvac
Nah, I put an elastic support around my head, just in case.Originally posted by condenseddave
Mid-
You are going to sprain your brain if you keep this up.

Using your calculator, solve the following:
x^3 = x * x * x = -27
or x = (-27)^(1/3)
A little figuring will show that 'x' must equal -3, i.e., -3 * -3 * -3 = -27
Why does the calculator fail to figure this out?
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01-18-2005, 10:57 PM #10Nope. Using apropriate math:Originally posted by midhvac
Hey! Wait a minute! Why are we using 1/6? Is it because the answer to b is 6? So I have to know the answer before I can work the problem?
a = (2 x 20^4) / b^6
a = 320,000 / b^6
b^6 = 320,000 / a
(b^6)^(1/6) = (320,000 / a)^(1/6)
b = (320,000 / a)^(1/6)
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01-19-2005, 03:58 AM #11
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I think I am going into another field. I will never figure this stuff out!!!
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01-19-2005, 05:19 PM #12
You big bunch of math nerds.
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01-19-2005, 05:49 PM #13Yea , bet they got pocket protectors too>)Originally posted by mattm
You big bunch of math nerds.


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