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Exponents

by masuarezdl » Thu Jan 29, 2009 10:26 am
The following problem is driving me crazy:

If (5^21) * (4^11) = (2 x 10^n), What is the value of n?
a. 21
b. 32
c. 64
d. XX
e. XX

Can anybody help out?
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by DanaJ » Thu Jan 29, 2009 10:40 am
(5^21) * (4^11 )= (5^21) *[(2^2)^11] = (5^21) * (2^22) = [(5*2)^21] * 2 = (10^21)*2, so n = 21.
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by masuarezdl » Thu Jan 29, 2009 10:58 am
DanaJ wrote:(5^21) * (4^11 )= (5^21) *[(2^2)^11] = (5^21) * (2^22) = [(5*2)^21] * 2 = (10^21)*2, so n = 21.
There we go! Thanks Dana.
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by ajmoney09 » Thu Jan 29, 2009 8:28 pm
The exponents have to be equal to multiple the base terms?
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by DanaJ » Thu Jan 29, 2009 10:44 pm
Yes, they sure do. Say you have (5^6)*(2^8). This is equal to (5^6)*(2^6)*(2^2) = (5*2)^6 *(2^2) = (10^6)*4. As you can see, we have 2^2 "on the side".
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by awilhelm » Fri Jan 30, 2009 7:52 am
My slightly slower approach:

5^21 * 4^11 = 2 * 10^n
5^21 * 2^22 = 2^1 * 2^n * 5^n
5^21 * 2^22 = 5^n * 2^n+1

From here you can see that n = 21
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The exponents

by vladmire » Fri Jan 30, 2009 8:59 am
ajmoney09 wrote:The exponents have to be equal to multiple the base terms?[/quote

I'm getting lost right here
(5^21) * (2^22) = [(5*2)^21] * 2
How is [(5*2)^21]*2 equal to (5^21) * (2^22)

The exponents aren't equal are they how can i multiply the base terms
Last edited by vladmire on Fri Jan 30, 2009 9:02 am, edited 1 time in total.
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Re: The exponents

by dmateer25 » Fri Jan 30, 2009 9:02 am
vladmire wrote:
ajmoney09 wrote:The exponents have to be equal to multiple the base terms?[/quote

I'm getting lost right here
(5^21) * (2^22) = [(5*2)^21] * 2
How is [(5*2)^21]*2 equal to (5^21) * (2^22)
Maybe this will help:

[(5*2)^21]*2 = 5^21 * 2^21 *2 = 5^21 * 2^22
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