Hi,

Can you please explain #37 in the Algebra book?

I'm trying to follow the answer in the back; however, it says you can factor out 3^x from 3^x-3^(x-1). and the result is 3x(1-3^-1). I think 3x is a typo and the result should be 3^x(1-3^-1). This has confused me and I can't follow the answer. Would you be so kind as to explain how to solve the problem here?

Thank you very much.

3^x - 3^(x-1) = ((3)^13)(2)

(3^(x-1))(3 - 1) = ((3)^13)(2)

(3^(x-1))(2) = ((3)^13)(2)

(3^(x-1)) = ((3)^13)

x - 1 = 13

x = 14

-- Veritas Help

How did you go from the first step to the second?

Agreed - you've just restated what the answer book has. Can you please explain your approach?

I understand how to solve this by factoring out just the 3^x but factoring out the 3^x-1 isn't making sense to me.

how did you get this: 3^x-1(3-1)??

Hey guys,

Great point...it's definitely not intuitive to factor out the 3^(x-1) term, and that's why this is a tough question.

One way to approach a problem like this, one with kind of vague numbers, is to do a quick "parallel" problem using small numbers.

So, here, the left hand side of the equation is:

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

Well, that's awkward...it's hard-to-visualize exponents. But try with small numbers to see the relationship and you'll better understand the concept. If x were to be 4, then

3^4 - 3^3 =

Well, here, it's clearer that the second term is common with the first. You could factor out 3^3 from both terms to get:

3^3 (3^1 - 1)

3^3 * 2

Seeing this, it may be clearer that the 3^x term is really the 3^(x-1) term just multiplied by one more 3. So you can factor out the smaller number, 3^(x-1), because of that.

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Or, you can rephrase 3^x in the original equation as: 3^(x-1) * 3^1

The reason you'd do that is because you want to find common terms to factor out, and by using the rule that x^y * x^z = x^(y+z), you can break apart x into x - 1 and 1 to get that common term.

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Either way, that's how you can factor out the common 3^(x-1) term as done in the explanation above...