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3^x - 2*3^(1-x) = 1 (pg. 3)
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| DJ Cinos |
| quote: | Originally posted by piggy
Divide both sides by 3^(1-x),
this gives you 3^(2x-1) - 2 = 3^(x-1).
Rearrange to get (1/3)*3^2x - (1/3)*3^x - 2 = 0
Let y = 3^x, then solve as a regular quadratic equation.
This gives you y = -2, 3
3 = 3^x , so x = 1. |
Right. Thanks for the explanation. :) I still have trouble getting it, but that's just my own stupidity. :p |
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| Cloudburst |
Mathematica FTW... :p
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| Jocker |
| quote: | Originally posted by piggy
Divide both sides by 3^(1-x),
this gives you 3^(2x-1) - 2 = 3^(x-1).
Rearrange to get (1/3)*3^2x - (1/3)*3^x - 2 = 0
Let y = 3^x, then solve as a regular quadratic equation.
This gives you y = -2, 3
3 = 3^x , so x = 1. |
correct, but you also have the second root, -2 = 3^x, solution to which involves complex numbers. |
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| yujie__ |
| roundhouse kick it |
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| ivanbee |
| the square root of an isosceles triangle is equal to the sum of it's sides |
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| plastikE |
log3^x - log6^(1-x) = log 1
x * log3 - (1-x) * log6 = log1
.47712x - .778 - .778x = 0
-.30088x = .778
x = -2.2587
? |
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| Mag1k |
| Just out of interest what grade is this for you? |
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| PaperBag831 |
u + me = us.
:clown: :clown: :clown: |
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| Cloudburst |
| quote: | Originally posted by Jocker
correct, but you also have the second root, -2 = 3^x, solution to which involves complex numbers. |
| quote: | Originally posted by Cloudburst
Mathematica FTW... :p
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:p |
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