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polynomial

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by vladmire » Mon Dec 15, 2008 2:13 pm
Take note whats in bold how do I factor out 2^x-1 out of the left side of the equation?
here's the deal: when you factor a power out of a polynomial, which power do you factor out: the smallest common power, or the biggest common power?
that's right, the smallest one. if you have x^5 - x^2, you only factor out x^2, not x^5.
analogy:
on this problem, just factor out 2^(x - 2).
this gives
left hand side = [2^(x - 2)](2^2 - 1)
= [2^(x - 2)](3)
you can then cancel the 3's, leaving 2^(x - 2) = 2^13. therefore, x - 2 = 13, so x = 15.
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Source: — Problem Solving |

by California4jx » Mon Dec 15, 2008 4:14 pm
Problem: 2^x - 2^x-2 = 3 (2^13)

Solution:

2^x (1 - 2^-2) = 3 (2^13)
2^x (3/4) = 3 (2^13)
2^x (1/4) = 2^13
2^x-2 = 2^13

with common base, it becomes:

x - 2 = 13
x = 15
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Re: polynomial

by California4jx » Mon Dec 15, 2008 4:18 pm
vladmire wrote:Take note whats in bold how do I factor out 2^x-1 out of the left side of the equation?
Not sure I understood your main concern, but here is the detail explain how you take out common

2^x - 2^x-2

can be written as:

2^x - 2^x * 2^-2

Now take out the common: 2^x

2^x (1 - 2^-2)
2^x (1 - 1/4)
2^x (3/4)

the rest is solved above.
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