## Fractional Exponent Over Variable

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
Arkmer
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### Fractional Exponent Over Variable

Hi, I was taking an economics exam today and was faced with the following after a few other simplifications...

$.033=\frac{k^{\frac{2}{3}}}{k}$

I was having trouble solving for my capital-labor-ratio (k). Since we were allowed to use our graphing calculator, I put the function into Y= and scrolled until the function equaled .033 which was at 27.033. Later in that question I found that my consumption level was negative... which is wrong, but that's a question for a different place.

I would just like to know if there is a real way to solve for k instead of my very fake way of doing it.

I did find "http://www.purplemath.com/modules/exponent5.htm" in the lessons section that showed how to manipulate numbers under a fractional exponent, but it only manipulates it; it never solves for a variable. Our professor said this was all basic algebra, but somehow I feel as though this one was a little above "basic".

jg.allinsymbols
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### Re: Fractional Exponent Over Variable

Right. This is just basic algebra. Know rules or laws of exponents.

$(b^m)/(b^n)=b^(m-n)$
The typesetting in my posting seems to not work.
Another try:
$frac\{b^m}{b^n}=b^(m-n)$

m and n may be rational.

PURPLEMATH! YOUR TYPESETTING IS NOT WORKING.

Another look; another try; on Feb. 21, 2013
$\frac{b^m}{b^n}=b^{m-n}$
Administrator's next comment seems strange, but I will include \,=\,
$\frac{b^m}{b^n}\,=\,b^(m-n)$
Curly braces for the exponent on the rightside is a key.
Last edited by jg.allinsymbols on Fri Feb 22, 2013 2:54 am, edited 4 times in total.

stapel_eliz
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$(b^m)/(b^n)=b^(m-n)$
The typesetting in my posting seems to not work.
Another try:
$frac\{b^m}{b^n}=b^(m-n)$

PURPLEMATH! YOUR TYPESETTING IS NOT WORKING.
Check out the basic LaTeX guide here. I'm not sure what wasn't working as intended in the first line, but "Another try" is missing the back-slash before the "frac" tag (rather than after the tag). Inserting the back-slash (and using curly-braces on the power on the right-hand side of the equation) yields this:

. . . . .$\frac{b^m}{b^n}\, =\, b^{m-n}$

Arkmer
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### Re: Fractional Exponent Over Variable

Thanks jg, I don't remember seeing that equation before, but it does get me a step further.

Once I have
$=b^{(m-n)}$
Then I can solve the exponent, yes? My situation left me with
$(\frac{2}{3}-1)=-\frac{1}{3}$
$.033=k^{\frac{-1}{3}}$
After reading through the whole section I linked again, I could not find how to completely remove an exponent... Which was kind of my goal/dilemma.

So now I look to Logs? Specifically

Which I rather like as a gif file.
$\log_k{(.033)}=\frac{-1}{3}$
Following what I think I should in "http://www.purplemath.com/modules/logrules5.htm" it asks me to simply evaluate a simple log showing similar to the following...
$\log_k{(.033)}=\frac{\ln(.033)}{\ln(k)}$
I still don't feel closer; logs are something I was never overly good with, so the introduction of them here is rather daunting. Am I gone too far or not far enough?

I did get my test back; I missed a .1 while putting together my depreciation and savings functions together (to find the steady state of growth). So realistically, after redoing everything up into my current equation, I should change the .033 into .33; not a huge change, but worth mentioning. It would explain my negative consumption rate though, lol.

jg.allinsymbols
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### Re: Fractional Exponent Over Variable

Learning rules of exponents must occur before learning about logarithms. What is the complete exercise or problem description for your first given example with the (k^(2/3))/k ?

The given problem will not require use of logarithms. You could use logarithms but not needed.
(I will do this without tex tags for a while until I learn better control of the code).
(You'll have to write the text form into symbolism on paper to see fully).

0.033=(k^(2/3))/k=k^(2/3-1)=k^(-1/3)
$0.033=\frac{k^{2/3}}{k}=k^{2/3-1}=k^{-1/3}$
That right-hand side can be undone to obtain k. You can and must do the same thing to the left-hand side and you will have solved for k.

(0.033)^-3=(k^(-1/3))^-3=k^((-1/3)(-3))=k
$(0.033)^{-3}=(k^{-1/3})^{-3}=k^{(-1/3)(-3)}=k$

Arkmer
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### Re: Fractional Exponent Over Variable

I'm actually a little upset that I got so focused on one side of the equation that I forgot I could manipulate both sides at once... putting both sides to the negative power of three would certainly solve the equation, but the answer is far from what I got in my little mathematical endeavor. That comes out to 27826 and some odd decimal.

The full problem, as you asked for is as follows.
- Begin Question -
A country has the per-worker production function: $y=6k^{\frac{2}{3}}$ Where y is out put per worker and k is the capitol-labor ratio. The depreciation rate is .1 and the population growth rate is .1.

The total saving function is $S=.1Y$ Where S is total national saving and Y is total output.

A. What is the steady state value of the capitol-labor ratio?
B. What is the steady state value of output per worker?
C. What is the steady state value of consumption per worker?
- End Question -

Part A is solving for k; lower and upper case are different in that lower case is per capita and upper is the aggregate value. The depreciation rate and population growth rate fit together in the equation $(.1+.1)k$ and the steady state is where $i=(.1+.1)k$. $i=sy$ What is s? $s=\frac{S}{k}=.1\frac{Y}{k}$ Take some substitution for $6k^{\frac{2}{3}}=(.1+.1)k$ (this is what I started to simplify in my first post here).

I'll stop here briefly to say I messed up, and you probably caught it. There is a .1 missing from my substitutions... The equation I should have come up with was $.1(6k^{\frac{2}{3}})=(.1+.1)k$ Not that it makes a difference for the operation of what I was asking originally.

Anyway, I progress awkwardly and incorrectly as stated, but it just shows how much math my professor is willing to do while correcting tests.

I don't know how much more of my work you would like, but the next two parts are rather irrelevant as my question is about part A.

Part B is just plugging in the value of k into $y=6k^{\frac{2}{3}}$

Part C needed some more number flipping, but I felt confident in this one. My math reads $Y=yk=1460.86$ which is an irrelevant number now, so I'm going to act like it's not there. $(yk)x.1=S$
$\frac{S}{k}=s=5.4$
$sy=i=292.01$
$y-i=c=-237.97$
More irrelevant numbers due to my missed .1 in the beginning; I'm not overly willing to do the math for it as long as the process was right (which it seemed to be since the .1 was all I was marked off for), and this took me forever to type out as well.

If you do see any more flaws, feel free to let me know; should help me on my final.