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Factoring algebraic equations with exponents

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by GMATGuruNY » Wed Aug 03, 2011 1:16 pm
oxfordbound wrote:Hi All,

I understand most algebra problems that have exponents (equivocate the bases and solve). But I ran into a problem from the Veritas Algebra book and the feedback provided from the Veritas book/forums wasn't up to par.

I'd like some real step by step guidance on this problem:

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

the answer is 14. I've worked through the problem as it has been laid out in the book but I cannot understand the methodology entirely in order to apply it to another problem.

The book denotes that you can factor out a 3^x OR (more efficiently) factor out a 3^x-1.

3^x-1 (3-1) = 2(3^13)
3^x-1(2) = 2(3^13)
x-1 = 13
x = 14

I don't fully understand this logic, can someone please clarify the underlying math logic/rules applied that is going on here please?

Thanks in advance,

Oxford Bound
The easiest approach would be to plug in the answers, which the GMAT would provide.

Answer choice: x=15
3¹� - 3¹�ˉ¹ = 2 * 3¹³
3¹� - 3¹� = 2 * 3¹³
3¹�(3-1) = 2 * 3¹³
3¹� * 2 = 2 * 3¹³

The exponent on the left side is 1 more than the exponent on the right side.
Thus, we need to subtract 1 from x=15.
x = 15-1= 14.
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by MBA.Aspirant » Wed Aug 03, 2011 1:31 pm
3^x - 3^x-1 = 2(3^13)

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

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

3^(x+1)- 3^x = 6 (3^13)

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

3^x * 2 = 2 * 3^14 (remove 2s)

3^x = 3^14

x = 14
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by Taniuca » Wed Aug 03, 2011 3:13 pm
I'm not an expert, but there is a simple way for solving this type of exponent equations, I use the common factor

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

3^x (1- 3^-1)= 2(3^13)-----------factorizing 3^x and remember that 3^-1= 1/3

3^x (2/3)= 2(3^13)

3^x=2(3^13)*(3/2)----------simplifying

3^x=3^14
so x=14
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by gmatboost » Wed Aug 03, 2011 9:54 pm
Hi,

I won't explain the whole question again, but I want to give you some background that hopefully will be helpful.

The exponent rule that everyone is using here is (a^m)(a^n) = a^(m+n)
And the corresponding rule is that (a^m)/(a^n) = a^(m-n)

So, when you see 3^(x-1), you can think of that (using the second rule) as (3^x)/(3^1) or just (3^x)/3 since 3^1 = 3.
And remember that dividing by 3 is the same as multiplying by 1/3, so (3^x)/3 = (1/3)*(3^x). You don't need to write it that way, but it can make factoring easier in the next step.

So, the equation is now 3^x - (1/3)*(3^x) = 2(3^13)

From there, you can factor out 3^x from both terms on the left to get (3^x)(1 - 1/3) = (3^x)(2/3).

At this point, the equation is (3^x)(2/3) = 2(3^13). Try solving from this point on your own and see how it goes.

Also, for future questions, if the left-hand side had said something like 3^(x+2) + 3^x, you could have done something similar using the exponent rule:
3^(x+2) + 3^x = (3^x)(3^2) + 3^x = (3^x)(9) + 3^x = (3^x)*(9 + 1) = (3^x)*(10)
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by jhoncorner » Wed Nov 09, 2011 10:18 pm
Algebraic Equation

A certain 3 digit number when divided by :a) 7 gives a remainder of 2. b) by 8 ...remainder of 3 and c) by 11.....remainder of 6.
I need the algebraic equation(s) to solve to get the 3 digit number.
Thanks
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by Anurag@Gurome » Wed Nov 09, 2011 11:31 pm
jhoncorner wrote:A certain 3 digit number when divided by :a) 7 gives a remainder of 2. b) by 8 ...remainder of 3 and c) by 11.....remainder of 6.
Say, the number is N.
Dividing N by 7, 8, or 11 gives a remainder of 2, 3, or 6.
Hence, (N + 5) is divisible by 7, 8, and 11.

Thus, (N + 5) must be a multiple of 7*8*11 = 616
Thus, possible values of (N + 5) are : 616, 1232, etc
Thus, possible values of N are : 611, 1227, etc

As N must be a 3-digit number, N must be 611.
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