integers and number properties

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integers and number properties

by jzebra10 » Sun Nov 27, 2011 6:18 am
How would you solve this?

if 5^11 x 4^5 = 5 x 10^k, what is the value of k?

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by Neo Anderson » Sun Nov 27, 2011 7:37 am
5^11 x 4^5 = 5 x 10^k
this can also be written as 5 x 5^10 x 4^5
or 5 x 5^10 x (2^2)^5
or 5 x 5^10 x 2^10
or 5 x (5 x 2)^10
or 5 x 10^10
thus k=10

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by tpr-becky » Mon Nov 28, 2011 11:20 am
realize that you can multiply distinct bases that have the same exponent together and get a new number that has the same exponent - for example - 2^2(3^2) = 6^2

Now to get to a 10 as a base you need a 5 and a 2 - you can work with the first part of the equation or the second - since Neo worked with the first, I'll show you how to work with the second.
5 x 10^k = 5 x 5^k x 2^k = 5^(k+1) x 4^k = 5^(k+1) x 4^(k/2)

On the other side we have 5^11 x 4^5 so we know k+1= 11 and k/2=5 thus for each k = 10.

The skill involved here is factoring and multiplying bases
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by Anurag@Gurome » Mon Nov 28, 2011 8:43 pm
jzebra10 wrote:How would you solve this?

if 5^11 x 4^5 = 5 x 10^k, what is the value of k?
Given: 5^11 x 4^5 = 5 x 10^k
Now 5 x 10^k = 5 x (2 * 5)^k = 2^k * 5^(k + 1)
Also, 4^5 = (2²)^5 = 2^10
So, 2^10 * 5^11 = 2^k * 5^(k + 1)
Now it can be seen that bases are the same on both sides, so equating the exponents, we get k = 10
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