yep
2006-10-12 16:55:18
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answer #1
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answered by Anonymous
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It's undefined if you're only talking about real numbers. No matter how much you try, you can't raise a number to a real power and get a negative result.
If you have a TI 83+ and set the mode to a+bi, ln(-1) will not return undefined but (pi)*i, an imaginary solution. If you want to know why, it's because of Euler's formula/identity: e^(ix) = cos(x) + i*sin(x).
2006-10-12 17:03:06
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answer #2
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answered by Anonymous
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Basically what people above are saying about imaginary number are correct, however the log function is infact periodic in the imaginary domain so therefor there are infact an infinte number of solutions in the complex domain to ln of a negative number as adding 2pi*sqrt(-1) to an exponential does nothing
2006-10-12 20:43:08
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answer #3
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answered by Aaron C 1
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Yes, if you're restricting yourself to real numbers.
Remember, y = ln x is the inverse function of x = e^y. For any real number y, x = e^y > 0. Therefore, for any negative number x, it is not possible to find a y = ln x such that e^y = e^(ln x) = x.
2006-10-12 16:58:48
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answer #4
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answered by James L 5
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ln a = x
There is no way to get a negative a by varying the value of x
Rearrange the equation,
exp(x) = a
when (x) varies from -infinity to infinity,
the value of (a) goes from 0 to infinity.
So, you cannot find an answer for ln (negative integer) and hence it is undefined.
2006-10-12 17:11:44
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answer #5
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answered by ideaquest 7
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Yes. You could have negative results, but never negative integers inside the Ln or Log...
Log (+) = + or -
same with Ln...
2006-10-12 17:13:59
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answer #6
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answered by Adriana 5
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Yes and no. ln(-1) = Ï*sqrt(-1), but that's only if you allow complex numbers. In the real domain, logs of zero and negative numbers are not defined.
2006-10-12 17:02:07
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answer #7
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answered by Engineer-Poet 7
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Yes.
You can use a graphing calculator and graph y=ln (x) to see this.
Remember that (if a and b are numbers) ln (a) = b is saying that the number e raised to the power of b is equal to a. There is no power you can raise a positive number (like e) to to get a negative.
2006-10-12 17:07:04
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answer #8
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answered by iMi 4
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It's not a real number, but it's define in the set of complex numbers. FDor example, ln(-1) = pi* i, because e^(pi*i) = cos(pi) + i*sen(pi) = -1 + 0 = -1
2006-10-12 16:58:49
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answer #9
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answered by Steiner 7
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yes
ln a = x means e raised to x gives 'a'.
since e is positive, 'a' will always be positive (even if x is negative).
so you can't have the ln of a negative number.
2006-10-12 16:56:23
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answer #10
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answered by Anonymous
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in the real domain it is undefined, but in complex domain it is defined.
for example
e^pi i = -1
so ln(-1) = i* pi
so ln x = ln (-x) + i* pi
2006-10-12 17:02:51
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answer #11
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answered by Mein Hoon Na 7
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