Mathematica FTW...

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Cheater!

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use the force.

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WRONG!

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Last edited by Grrrrr on Feb-12-2006 at 23:21

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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roundhouse kick it

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

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

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

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

-.30088x = .778

x = -2.2587

?

Just out of interest what grade is this for you?

u + me = us.

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...

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