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DS OG # 131 factoring exponents

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Re: DS OG # 131 factoring exponents

by nitya34 » Mon Jul 27, 2009 7:50 am
you yrself have done it

taking 5^k from LHS
whats yr doubt?
sudi760mba wrote:I'm a little confused on factoring exponents.

5^k - 5^k(5^-1)=500
next
5^k-5^k(1/5) = 500
When we factor 5^k, how do we then get:
5^k(1-1/5)?

Thanks!
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by sudi760mba » Mon Jul 27, 2009 8:02 am
well exactly this:

5^k-5^k(1/5) = 500

how do we go from what we have above to what we have below?

5^k(1-1/5)?

I mean when you factor 5^k wouldn't be 5^k(1-1(1))?
5^k(1/5)= 1?

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by navalpike » Fri Jul 31, 2009 9:20 am
So you are trying to figure out how to go from 5^k-5^k(1/5) to 5^k(1-1/5)?
We simply take 5^k as a common factor
5^k
Then we multiply this by 1 to get the original expression of 5^k
5^k(1
Now we need to get the second part of the expression, which is -5^k(1/5)
Since we already took out 5^k as a factor, no need to repeat it. We get…
5^k(1-1/5)
It is actually more intuitive when you see this backwards. Multiply 5^k with (1-1/5), first by one and then by – 1/5, and it gives you 5^k – 5^k(1/5)
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