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 Post subject: Lesson 4, Algebra Question 37Posted: Sat Jul 30, 2011 3:09 pm

Joined: Fri Jun 24, 2011 2:30 pm
Posts: 3
Question 37 in Lsn 4, Algebra. I do not understand the book answer.

How does 3^x - 3^(x-1) factor into

3^x ( 1 - 3^-1) or

3^(x-1) ( 3 - 1 ). I understand how it multiplied by 1 gives itself. But when 3^x-1 is multiplied by 3, it gives the original 3^x from the first equation. I can't seem to grasp this. Thanks.

Dustin Healey

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 Post subject: Re: Lesson 4, Algebra Question 37Posted: Fri Aug 05, 2011 5:16 pm

Joined: Thu Jul 03, 2008 2:13 pm
Posts: 117
Hey Dustin,

Where they make this tricky is by including the subtraction in the second exponent. But say it were:

3^(y+1) - 3^y

You'd likely see that 3^(y+1) = 3^y * 3^1, so you'd factor the common term in:

3^y * 3^1 - 3^y

to get

3^y (3 - 1)

Right?

Well, the key here is that the subtraction of exponents means that in the given problem:

3^x - 3^(x-1)

You can't factor out a common 3^x because the second term doesn't have a full x...it's one short. But you can express that as:

3^x - 3^x * 3^(-1)

That allows you to use that common x:

3^x (1 - 3^(-1))

And proceed from there.

In either case, the key is recognizing that an exponent that contains either addition or subtraction can be expressed as two different same-base-different-exponent terms multiplied together:

x^(y + z) = x^y * x^z

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