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Decimal Exponents

Date: Wed, 02 Nov 1994 14:36:16 -0800
From: Caryl Lynn Segal
Subject: problem

economics textbook.  This involves food production and the equation is:


K involves machines and L involves labor.

The diagram shows Q(output) as 13,800
                              L as 760
                              K as  90

numbers.  Any help you can provide will be appreciated.

P.S. I know this is set up for K-12 but am hoping you are not too busy to 
help out someone beyond this level

Date: Wed, 2 Nov 1994 15:40:28 -0500
From: Melissa D. Binde
Subject: Re: problem

I had trouble too--is there any more information you can give us?  What
kind of economics (micro, macro, etc.) is this?  I'm not an expert in
economics, but perhaps some of the other "doctors" will be able to help you
later this evening.

>help out someone beyond this level

We'll do our best!


From: Dr. Ethan
Subject: Re: problem
Date: Wed, 2 Nov 1994 15:50:06 -0500 (EST)

760 to the .8 power times 90 to the .2 power times 100 equals 13789.8, 
which rounds to 13800 so I think we are just dealing with multiplication 
and exponents.
                Ethan Doctor-On Call

Date: Wed, 02 Nov 1994 17:24:42 -0800
From: Caryl Lynn Segal
Subject: Re: problem

I guess what I'm trying to discover but didn't explain is how you get from 
K.8 power to learn it is 90. My problem is being unable to figure out how  
to work with .8 power or .2 power etc.

Date: Wed, 2 Nov 1994 18:28:44 -0500
From: Gabe Farmboy Cavallari
Subject: decimal exponents

Here's a possible explanation.  A decimal exponent could also be thought of 
as a fractional exponent, so .8 would equal 4/5.  The fractional expression, at
least for me, is easier to use, because raising a number to the 4/5's power
is like raising that number to the 4th power and then taking the 5th root.
For any number raised to the m/nth power, raise the number to the mth (from
the numerator)power and take the nth (from the denominator) root.

As a more familar example, consider the square root.  The exponential
equivalent would be to raise a number to the 1/2 power, or equivalently
the .5 power.

Hope this helps.

Gabe Cavallari, math doc

Date: Wed, 2 Nov 1994 18:34:21 -0500
From: Heather Stickney
Subject: Re: problem

     From what you've told us, you can't do the problem.  What I'm thinking is
that this equation is perhaps a standard equation to use to determine
output when you know K and L, or to determine L when you know K and 
how much output you want, or vice-versa.  I think the diagram is just showing
an example of how they used the equation.  When K=90 and L=760, there is 
an output of 13,800.  You are not expected to use the equation to determine
the values of all the variables at the same time. If this doesn't make
sense, or I've misunderstood what you are asking, write back.>

Date: Wed, 2 Nov 1994 22:46:21 -0500 (EST)
From: "Michael W. S. Morton"

     I think there may be another way of looking at this problem (that may be 
more in depth than you care to go, or even want to think about.   :)  Having 
taken a little econ, I know there are such graphs as isoquants, that is, 
graphs where the output of a firm stays constant as you move along the 
curve, varying K and L.  These curves show the relationship between 
capital (K) and labor (L) at different points, holding the output (Q) 
constant.  So, say.....

        Q = 13800
     The function then looks like:
                       .8    .2
        13800 = 100 * K   * L

               .8    .2
        138 = K   * L

         .8         .2
        K  = 138 / L
     This function can be graphed (most preferably by a computer).  It 
represents the way in which capital and labor can be interchanged at a 
constant output of 13800 units.  The slope of this line at any given 
point is equal to the Marginal Rate of Technical Substitution (MRTS) at that 
point.  More curves can be put on this graph by letting Q = different values.

     This may be going further than you wanted to, or maybe it helps.  
If you have any questions, just lemme know!

                                        -MORTON, A math Doctor
                                                 w/a little econ :)
Associated Topics:
High School Exponents

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