I just want to make sure I am understanding this corectly...for some reason factoring these particular scenarios gives me problems.
Anyways-stolen from the GmatPrep stuff:
3^x - 3^x-1 = 162
It seems there are two ways to factor this one through:
1. 3^x (1-(1/3)) = 162
2. 3^x-1 ((3^1)-1) = 162
Now, to make sure I am understanding these correctly, if it was say:
3^x - 3^x-2 = xxx (this number is a non-issue for me)
Would the two ways to factor become:
1. 3^x (1-(1/9)) = xxx
2. 3^x-2 ((3^2)-1) = xxx
I just want to make sure I am understanding it correctly. I guess, in the same manner, you could take 5^x-1 - 5^x-3 and turn it into:
5^x-3 ((5^2) - 1)
If I am wrong please let me know! Thank you!
Question about factoring (should be simple)
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Ur Absolutely right !!!uncbeers wrote:I just want to make sure I am understanding this corectly...for some reason factoring these particular scenarios gives me problems.
Anyways-stolen from the GmatPrep stuff:
3^x - 3^x-1 = 162
It seems there are two ways to factor this one through:
1. 3^x (1-(1/3)) = 162
2. 3^x-1 ((3^1)-1) = 162
Now, to make sure I am understanding these correctly, if it was say:
3^x - 3^x-2 = xxx (this number is a non-issue for me)
Would the two ways to factor become:
1. 3^x (1-(1/9)) = xxx
2. 3^x-2 ((3^2)-1) = xxx
I just want to make sure I am understanding it correctly. I guess, in the same manner, you could take 5^x-1 - 5^x-3 and turn it into:
5^x-3 ((5^2) - 1)
If I am wrong please let me know! Thank you!
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Yes, with that one small correction, everything looks perfect.
3^x - 3^(x-1) = 7[ (3^x/7) - (3^(x-1)/7)]
or you could factor out 3^2:
3^x - 3^(x-1) = (3^2)[3^(x-2) - 3^(x-3)]
The main thing is to recognize what would be most useful to factor out, and in the question
3^x - 3^(x-1) = 162
the most useful factorizations are the two that you suggest: factoring out 3^x, or probably even better, factoring out 3^(x-1).
I'd just add one thing- you can factor 3^x - 3^x-1 in more than two ways, though most of the ways you might factor won't be helpful. You can factor it any way you like, in fact. You could factor out a 7 if you want to:3^x - 3^x-1 = 162
It seems there are two ways to factor this one through:
1. 3^x (1-(1/3)) = 162
2. 3^x-1 ((3^1)-1) = 162
3^x - 3^(x-1) = 7[ (3^x/7) - (3^(x-1)/7)]
or you could factor out 3^2:
3^x - 3^(x-1) = (3^2)[3^(x-2) - 3^(x-3)]
The main thing is to recognize what would be most useful to factor out, and in the question
3^x - 3^(x-1) = 162
the most useful factorizations are the two that you suggest: factoring out 3^x, or probably even better, factoring out 3^(x-1).