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

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