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Exponents

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Exponents

by vladmire » Fri Jan 30, 2009 10:08 am
I was wondering if I come across a problem that includes several integers with exponents what is the quickest way to solve for example

(13^8)(26^3)(10^43)(8^22)(7^14)(19^30)

is there a quick way to figure problems like these out? OR would I not come across something like this on the gmat.I don't have a great formula to solve exponents except for multiplying each number very time consuming..
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by masuarezdl » Fri Jan 30, 2009 11:23 am
I believe that what you have o do is to use the laws of exponents by factoring out.

For example, if you have 4^2 * 2^2, then you can simply specify that as 2^2*2^2*2^2 = 2^6.

I had a problem as follows, (solved by Dana):

(5^21)*(4^11) = 2x10^n, what is the value of n?

(5^21)*(2^11)*(2^11) = 2x10^n ---> Factor the 4^11
(5^21)*(2^22) = 2x10^n
(5^21)*(2^21)*(2) = 2x10^n
(10^21)*(2) = 2x10^n
2x10^21 = 2x10^n

Therefore, n = 21.
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by DanaJ » Fri Jan 30, 2009 11:26 am
The best way to approach this is by doing the old but trusty factorization. You get that:
13 = 13, so 13^8 = 13^8
26 = 2 * 13, so 26^3 = (2^3)*(13^3)
10 = 2*5, so 10^43 = (2^43)*(5^43)
8 = 2^3, so 8^22 = 2^66
7 = 7, so 7^14 = 7^14
19 = 19, so 19^30 = 19^30.

Now we add all the powers of the factors and we get:
(2^112)*(5^43)*(7^14)*(13^11)*(19^30)
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