Fractions and Powers again?

Question

5K^(-1/2) C^(1/2)-L=0
5K^(1/2) C^(-1/2)+-4L=0
K+4C=16

This is part of a bigger problem on lagrangean functions - I need to know how I can solve for C, K and L. My main problem is that I do not know how to deal with the powers. I multiply the first equation by 4 then want to take the second to eliminate L but how do I deal with the powers?

What would the rules be regarding something like:

[C^(-2/3) B^(2/3)] / [C^(1/3) B^(-1/3)]

Powers confuse me! - I have an exam in this s*** tomorrow :(

No sign between K and C - if there was I think I could solve it.

Ok...So what if the powers aren't the same on the top and bottom of the fraction?

Answer 1

c^(1/2) is the same as saying C^(0/5) or square root of C and B^(2/3) is the same as saying 2/3 root of B or
B raised to the 0.66667 power B^0.6667 in decimal form.

[C^(-2/3) *B*(2/3)] / [C^(1/3) * B^(-1/3)]

divide C^(-2/3) by C^(1/3) which gives you C/C^(-2/3 - 1/3)
this gives you 1^(-1) i.e. (-2/3 +(- 1/3)) = -1
1^(-1) = 1
so now you have after simplifying and clearing C:

B^(2/3) / B^(-1/3) ok! that gives you (B/B)^ (2/3 - (-1/3))
B divided by B = 1

(2/3 - (-1/3)) is going to equal to (2/3 + (1/3)) which = 1

so you have now 1^1 which equals to a 1

so you answer in this case is 1

[C^(-2/3) * B^(2/3)] / [C^(1/3) * B^(-1/3)] = 1

Always keep in mind when dividing powers exponents you simply change the sign of the power in the divisor and add. It is called additive inverse. Fancy name for substracting signed numbers.
When you are multiplying the powers exponents you add straight forward (not changing of signs)

When adding and substracting numbers with powers (exponents) it is like adding and subtracting fractions the exponents (powers) have to be the same or common.
Examples:

A^2 + A^2 = 2A^2 you can added them because they have a common power (exponent)
example: let A = 5
5^2 + 5^2 = 2(5^2)
25 + 25 will be 50 and 2* 5^2 will be 2* 25 which will be 50 as well so
5^2 + 5^2 =2(5^2)
50 = 50

Now if you have say
A^2 + B^2 =?
then the answer for that in simpliest terms is going to be:
A^2 + B^2 = A^2 + B^2 that is the simpliest term and answer because:
Example: we will let A = 5 and B=2
in this problem the answer will be.
5^2 + B^2 =
25 + 4 = 29 correct.
But if we try to simplify it first it turns out like this:

5^2 + 2^2 = 10^2
25 + 4 = 100
25 +4 = 100 it turns out:
29 = 100 which is definitely not the right answer because they are nowhere near equal.

A^3 + A^2 = A^3 + A^2 because the numbers do not have the same power (exponents) you can not add them together. you will have to clear your exponents first before adding the two numbers together.

Also you can add numbers with exponents where the numbers are not equal.
example:
A^3 + A^2 = ? you would think the you can simplify it to become 2A^5 but you can not.
example:
we will let A = 5

try working it where A^3 + A^2 = 2A^5 first.

5^3 + 5^2 = 2(5^5)
125 + 25 = 2(3125)
125 +25 = 6250
150 = 6250 definitely no equal. so A^3 + A^2 = 2A^5 is definitely wrong in this case you can only do it this way:

A^3 + A^2 =?
5^3 + 5 ^2 = ?
125 + 25 = 50

Hope that helped you a little. The key is not to get confused just take it one step at a time. And not short cuts until you get comfortable with it. Show each step no skipping steps because if you will not know which step you did wrong if you do not show each step. And most important nor will your teacher.
I know when I was stuck teaching Trig and Algebra on a 25 problem test or work assignment I would usually assign 3 points for the work and 1 point for the correct answer. And maybe a 1/2 point off if they worked the problem correctly but just got a sign off. I was more interested in them being able to work the problems out correctly. Reason I was going to get them later in engineering classes.

Always keep this in mind when work Algebra, Trig or calculus. The letters represent real numbers that you are substituting the numbers with:
Reason so you can get a complex problem simplified down as much as possible with out having to deal with large numbers:
Then once you get the problem as simple as you can you plug the numbers in and work the problem.(equation) It is a lot easier dealing with A's B's C's X's and Y's than having to deal with 566000, 1200, 85467 etc when working out an equation.

Answer 2

OK. It may be better to read my answer to your other question first, where these first steps will be explained...

5K^(-1/2) C^(1/2) = √C ÷ 5√K = √(C/25K)

5K^(1/2) C^(-1/2) = 5√K ÷ √C = √(25K/C)

Rearranging your second equation gives:
√(25K/C) = 4L
If we square both sides, we get:
25K/C = 16L²

Similarly for the first equation:
√(C/25K) = L
squaring both sides gives:
C/25K = L²
and multiplying by 16 gives:
16C/25K = 16L²

Now we have:
25K/C = 16C/25K
multipling both sides by C gives:
25K = 16C²/25K
and then multiplying by 25K gives:
625K² = 16C²
or 25K = ±4C

substituting K = ±4C/25 into your third equation:
K = 16 - 4C = ±4C/25

add 4c to both sides:
16 = 104C/25 or 96C/25
C = 400/104 or 400/96

But one of these values will give a negative K, and we'll be taking square roots. If you want a solution involving imaginary numbers you'll have to pay ..... someone else!

Putting C = 400/104 into K=16-4C gives K=64/104
Substituting these values into either of the original equations gives L = 52/104 :that's a half.

Points to remember:
1/a ÷ 1/b is the same as b / a
√a ÷ √b is the same as √(a/b)
when multiplying numbers, add the powers
when dividing numbers, sutract the powers

I hope the exam went well!

Answer 3

a negative power is the same as a power in the denominator, so C^(-2/3)B^(2/3) is the same as B^(2/3) / C^(2/3).

So B^(2/3)C^(-2/3) would be equal to B/C, and since the powers are the same in the numerator and denominator, they cancel and leave only the B/C terms.

do the same for the bottom and divide out the whole thing. after doing this, I get B^2/C^2 as the final answer for the bottom problem.

Answer 4

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Answer 5

is there supposed to be a plus or minus sign in these between k and c?

Answer 6

Apologoes I need points.xx