Need help with math problem

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• 01-18-2005, 07:03 PM
midhvac
In this formula b = 6

a = (2 x 20^4) / b^6

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?

• 01-18-2005, 07:22 PM
coolwhip
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!

• 01-18-2005, 08:31 PM
condenseddave
Mid-

You are going to sprain your brain if you keep this up.:D
• 01-18-2005, 08:38 PM
westcoast refer man
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?
• 01-18-2005, 10:53 PM
Andy Schoen
Quote:

Originally posted by midhvac
In this formula b = 6

a = (2 x 20^4) / b^6

Perhaps some number theory may be appropriate here.

'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.

Quote:

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?

Ok, lets assume 'a' = 7, what is 'b'?

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? :confused:
• 01-18-2005, 11:17 PM
midhvac
You guys rock! Thanks! I've got a scientific calculator and the 1/6 power works.
• 01-18-2005, 11:19 PM
midhvac
Quote:

Originally posted by condenseddave
Mid-

You are going to sprain your brain if you keep this up.:D

Nah, I put an elastic support around my head, just in case.
• 01-18-2005, 11:46 PM
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?
• 01-18-2005, 11:50 PM
Andy Schoen
Quote:

Originally posted by midhvac
Quote:

Originally posted by condenseddave
Mid-

You are going to sprain your brain if you keep this up.:D

Nah, I put an elastic support around my head, just in case.

Let's put that elastic support to work... :)

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? :)

• 01-18-2005, 11:57 PM
Andy Schoen
Quote:

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?

Nope. Using apropriate math:

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)
• 01-19-2005, 04:58 AM
ac rookie
I think I am going into another field. I will never figure this stuff out!!!
• 01-19-2005, 06:19 PM
mattm
You big bunch of math nerds. :D :D
• 01-19-2005, 06:49 PM
glennwith2ns
Quote:

Originally posted by mattm
You big bunch of math nerds. :D :D
Yea , bet they got pocket protectors too>)
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