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-- 3^x - 2*3^(1-x) = 1
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Posted by Cloudburst on Feb-12-2006 20:13:

Mathematica FTW...

Posted by sensorium on Feb-12-2006 20:15:

Cheater!

Posted by kadomony on Feb-12-2006 23:03:

use the force.

Posted by Grrrrr on Feb-12-2006 23:10:

WRONG!

Posted by Jocker on Feb-12-2006 23:15:

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.

Posted by yujie__ on Feb-13-2006 02:18:

roundhouse kick it

Posted by ivanbee on Feb-13-2006 02:24:

the square root of an isosceles triangle is equal to the sum of it's sides

Posted by plastikE on Feb-13-2006 04:10:

log3^x - log6^(1-x) = log 1

x * log3 - (1-x) * log6 = log1

.47712x - .778 - .778x = 0

-.30088x = .778

x = -2.2587

?

Posted by Mag1k on Feb-13-2006 04:45:

Just out of interest what grade is this for you?

Posted by PaperBag831 on Feb-13-2006 04:55:

u + me = us.

Posted by Cloudburst on Feb-13-2006 09:57:

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...

Posted by Nabistai on Feb-13-2006 14:00:

quote:
Originally posted by plastikE
log3^x - log6^(1-x) = log 1

x * log3 - (1-x) * log6 = log1

.47712x - .778 - .778x = 0

-.30088x = .778

x = -2.2587

?

Indeed, just use log. It's extremely simple really

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