# Homework Help: Basic question about exponents

1. Sep 1, 2011

### rygza

so i need to find a,b, and c:

(M^0)*(L^0)*(T^1) = (M^a+b+c)*(L^-a-3b+2c)*(T^-2a-2c)

to solve i equated the exponents to that of each matching variable:

a+b+c = 0

-a-3b+2c = 0

-2a-2c = 1

Is there a mathematical process of obtaining the above equations besides simply "equating" the exponents? Like taking the natural log or something like that?

Last edited: Sep 1, 2011
2. Sep 1, 2011

### Ray Vickson

When you write M^a+b+c, this means a + b + M^a when read according to *standard mathematical rules*. If you really mean M^(a+b+c), you need to use parentheses (as I have just done), or else typeset using LaTeX or use the "superscript" commands.

To answer your question: whether of not you have the equations for a, b and c that you wrote (and one of them is wrong: it should be a+b+c=0), depends on whether or not the original equation is supposed to hold for _all_ M, etc. If the M, N, etc. are just some given numbers then there may be infinitely many different solutions. For example, if I have an equation of the form M^0*N^1 = M^(a+b) * N^(2a-b) and I want it to hold for all M and N, then I do need a+b=0 and 2a-b = 1, so a = 1/3 and b = -1/3. However, if, for example, I have just some fixed values such as N = 2 and N = 3 then my equation says 3 = 2^(a+b) * 3^(2a-b) and any solution has the form a = .3800937668+.1402813002*b, so b can be anything. And, yes, in this case I just used logarithms to get the solution (or, rather, the computer algebra system Maple 11 did).

RGV

3. Sep 1, 2011

### rygza

You obviously understood what i meant. yes i know i didn't express everything in standard form and i did make a mistake with the first equation (now fixed). I didn't see the need, as equating the exponents would make it clear what i meant (which it seems it did to you).

Regarding my actual question, thanks but you weren't very straight forward in addressing it

4. Sep 1, 2011

### Ray Vickson

Well, *you* still have not addressed the issue of whether or not the equation was supposed to hold for all M, N, etc., or for just some given values. That makes all the difference in the world, and I did address it.

RGBV

5. Sep 2, 2011

### rygza

I thought it was obvious that it does hold for all M,L, & T. If there were fixed values for them I would have included that information haha. You addressed the question but I stated that it wasn't straight forward. I already know what to do if there were fixed values (that's why I asked about logarithms), I was simply asking for alternative methods to solve *this* specific case.

6. Sep 2, 2011

### Ray Vickson

OK, so you want to be convinced that if M^p * L^q * T^r = M^u * L^v * T^w holds for all positive M, L and T then we necessarily have p=u, q=v and r=w. Actually, it is enough to assume this holds for all M, L and T in some non-empty intervals greater than zero, for both sides are then nonzero and so we can divide to get M^(p-u) *L^(q-v) *T^(r-w) = 1 for all M, L and T in the intervals. Fixing L and T we have that M^(p-u) = constant as M varies in an interval, so differentiating w.r.t. M gives 0 = (p-u)*M^(p-u-1) for all M, so p-u = 0.

We don't even need the equation to be true for all M, L and T in intervals; it is enough to have them true for at least two different values of each of M, L and T; in that case, however, you do need to take logarithms to get the result.

RGV

7. Sep 2, 2011

### Staff: Mentor

Yes, Ray understood what you meant, even though it didn't agree with what you actually wrote. As someone who is coming to PF for help, don't make it harder for us to fathom what you're asking.

When someone writes an expression such as a + b/c + d, we can interpret this the right way, as $a + \frac{b}{c} + d$, or as what we think the poster really means, $\frac{a + b}{c + d}$. Trying to decipher what someone writes takes time that could be better spent on helping the poster.

Writing M^a+b+c falls into the same category.

8. Sep 2, 2011

### rygza

I actually did try dividing them but then didn't know what to do from there. Thanks, this was of much help