Dealing with exponent laws (2 simple questions in one thread)

1. Feb 6, 2008

mike_302

1. The problem statement, all variables and given/known data

6^1+6^-1 / 6^1-6^-1 (Question is to evaluate that, but I am going to venture to guess that we are supposed to somehow simplify the question a lot further. We just finished learning all the exponent laws)

Explain how you can tell which is bigger without evaluating: 20^100 or 400^20 ?

Have not been able to evaluate at all.

2. Feb 6, 2008

kuahji

x^-1 = 1/x

From there, you can finish it a number of ways. I'd probably just combine the fraction & then divide the whole thing to get a fraction result.

To find which is larger, they both start with 6, & either add or subtract x^-1. If you add a positive number to 6 or subtract a negative number, the positive number will be larger.

3. Feb 6, 2008

mike_302

OH! haha. The second part with "which is bigger" is a whole differnt question. Sorry I didn't make that clear :P .

Anyways, since only the fist question was answered with accuracy here, I would like to discus that quickly. How do you combine the fraction like you say? That is where I am getting mixed up: Rearranging to get all positive exponents.

4. Feb 6, 2008

HallsofIvy

Staff Emeritus
$$\frac{6+ 6^{-1}}{6- 6^{-1}}$$
Multiply both numerator and denominator by 6.

$20= 2^2(5)$ so $20^{100}= 2^{200}(5^{100})$. $400= 40(100)= (8*5)(4*25)= 2^5(5^3)$ so $400^{20}= 2^{100}(5^{60})$
Can you compare those?

5. Feb 6, 2008

mike_302

ahhh! Yes, I see for both now. I understand the first one and well... The second one, I get the idea of making them both have similar bases but how you did it would take a little more concentration on my behalf. I will do that after posting this but I must thank you for your work.

6. Feb 6, 2008

kbaumen

Actually, it is easier to solve if you see that $400 = 20^2$. from which you get that $400^{20} = 20^{2 * 20} = 20^{40}$ and of course $20^{100} > 20^{40}$.

EDIT: It's a property of exponents that $(a^b)^c = a^{bc}$

Last edited: Feb 6, 2008
7. Feb 6, 2008

mike_302

You're right. I understand that now, no problem. Thanks for your help as well! :)