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TheNewt
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Old 02-15-2007 , 21:44   Re: r
#1

you can't express i as a real number, i is the sq root of -1 O.o


Solve this Hawk:
ln(3x) = ln(5)^3
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Hawk552
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Old 02-15-2007 , 22:37   Re: r
#2

Quote:
Originally Posted by MysticDeath View Post
you can't express i as a real number, i is the sq root of -1 O.o


Solve this Hawk:
ln(3x) = ln(5)^3
ln(3x) = ln(5)^3
e^[ln(3x)] = e^[ln(5)^3]
// I'm too lazy to do it properly here
e^[ln(3x)] = e^(4.17...)
3x = e^4.17...
// calculator doesn't go past that, sorry
x = 21.54835061...
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FormulaZero
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Old 02-15-2007 , 22:14   Re: r
#3

Is the answer:

E = MC2?
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SweatyBanana
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Old 02-15-2007 , 22:27   Re: r
#4

q = mcΔT
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TheNewt
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Old 02-15-2007 , 22:36   Re: r
#5

Quote:
E = MC2?
Energy is equivalent to the mass times speed of light skvared!
3^(4x+5) = 9^(7x+21)
O.o
what is X?
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TheNewt
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Old 02-15-2007 , 22:42   Re: r
#6

-edited out this rude comment; I apologize hawk -
ln(3x) = ln(5)^3 is the same as saying
3x = 5*3
x=5

Ill explain step by step for the ones who aren't excelling in math.
ln(3x) = ln(5)^3 has two logarithmic functions on either side of the equation, if both functions have the same base, in this case the e becuase its the natural log. You can drop the log function, and any exponent on the logarithmic function is brought down to multiplication. (You can bring down the exponent first if you want)
ln(3x) = 3ln(5)
and then drop the ln from either side
3x = 3*5
devide 3
x = 5
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Hawk552
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Old 02-15-2007 , 22:56   Re: r
#7

no, it's not, you typed it wrong if you're basing it on that law

It would have to be:

ln(3x) = ln(3^5)

but you added the ^5 after it, thus your solution is wrong. I tried working it out based on your solution and got 2.70... = 4.169..., which is obviously wrong.
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stupok
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Old 02-15-2007 , 22:56   Re: r
#8

No, Hawk is right.

ln(3x) = ln(5) + ln(3) --> 3x = 5 * 3
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TheNewt
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Old 02-15-2007 , 22:58   Re: r
#9

mm Yeah I was thinking about that actually; couldn't find my notes on it, but I think your right. (as freaking usual... lol)

stupok: That has almost no relevance, we are talking about ln(5) to the power of 3 not ln(5 times 3) which would result in ln(5) + ln(3)

Okay I worked this out on paper:

ln(3x) = ln(5)^3
You CAN bring down the 3
ln(3x) = 3ln(5)
this is same as
ln(3x) = ln(5) + ln(5) + ln(5)
drop the natural log
3x = 5 + 5 + 5
3x = 15
x = 5
:/

My original method to solve it
ln(5)^3, is the same as ln5^3
Parenthesis have no relevance when there is nothing happening to the 5 inside the parenthesis


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Hawk552
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Old 02-15-2007 , 23:30   Re: r
#10

... I don't think you got my point. It's still wrong.

What you basically did is this:

ln(5)^3 = 3ln(5)

or, for a better example:

2^3 = 3*2

Is this correct? Not by any stretch of the imagination. Sub x into your original equation, you'll get a totally wrong answer.

Oh, just the kicker, I'm writing this out for people who "don't excel at math".
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