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    Okay, so I was solving a question and got down to ln((\frac{x^2}{3})=ln12. From that I can easily see that x = +- 6. However, the book says, that as whatever is in the brackets can not be less than or equal to 0, x is only = +6. But this doesn't make sense? x^2 negates the negative sign so it should work.

    Here's the full question: Solve, giving your answer(s) as exact values of x: 2lnx-ln3=ln12

    Now this brings me onto another point. Let's say we tried doing ln(-3). Obviously that can't be done. How about 2ln(-3)? Still can't be done. But by the power rule, isn't 2ln(-3)=ln((-3)^2)=ln(9), which can be worked out?
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    (Original post by ViralRiver)
    Now this brings me onto another point. Let's say we tried doing ln(-3). Obviously that can't be done. How about 2ln(-3)? Still can't be done. But by the power rule, isn't 2ln(-3)=ln((-3)^2)=ln(9), which can be worked out?
    I'm not actually sure about this, so hopefully someone will correct me if wrong.

    You can evaluate things like ln(-3), they just have a complex part:

    ln(-1) = i\pi (Easy to see from Euler's identity: e^{i\pi} +1 = 0, )

    so ln(-3) = ln(3) + ln(-1) = 3 + i\pi.

    Likewise, 2ln(-3) = 2ln(3) + 2ln(-1)= ln(9) + 2i\pi

    The real part of your answer is still as expected (ln(9)), but there's an added imaginary part.

    If you haven't come across complex numbers yet that might not make sense. If I'm honest I'm still not sure why we can't just say "well -3 squared is +9, let's ignore the complex part"!
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    (Original post by EEngWillow)
    I'm not actually sure about this, so hopefully someone will correct me if wrong.

    You can evaluate things like ln(-3), they just have a complex part:

    ln(-1) = i\pi (Easy to see from Euler's identity: e^{i\pi} +1 = 0, )

    so ln(-3) = ln(3) + ln(-1) = 3 + i\pi.

    Likewise, 2ln(-3) = 2ln(3) + 2ln(-1)= ln(9) + 2i\pi

    The real part of your answer is still as expected (ln(9)), but there's an added imaginary part.

    If you haven't come across complex numbers yet that might not make sense. If I'm honest I'm still not sure why we can't just say "well -3 squared is +9, let's ignore the complex part"!
    Ahhh.. I see, but that makes it insanely confusing for my C3 exam, as I can't use complex numbers there. So do I just have to say that 2ln(-3) doesn't work?
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    Yes, in C3 stick with real numbers, so you then can't have the log of a negative number or 0. Since the original question had ln x in it, x must be positive.
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    but its x^2 so it is always positive since negative^2 is positive you can have -6
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    (Original post by ViralRiver)
    Ahhh.. I see, but that makes it insanely confusing for my C3 exam, as I can't use complex numbers there. So do I just have to say that 2ln(-3) doesn't work?
    In C3, complex numbers don't exist. It'd be worth including modulus signs in your lns, i.e. ln|x|, and only dealing with the positive values. ( I think some of the mark schemes used to have this included, actually?)
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    But what I still don't understand, 2lnx equates to lnx^2, which is always positive, regardless of whether x itself is positive or not.
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    (Original post by ViralRiver)
    But what I still don't understand, 2lnx equates to lnx^2, which is always positive, regardless of whether x itself is positive or not.
    Log/ln cannot be negative simply in C3.

    2lnx = lnx^2 Because of log rules, you can bring down the power as a multiplied value
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    what's the answer for x? I got e^ln36/2
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    (Original post by Faith01)
    what's the answer for x? I got e^ln36/2
    x = +6
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    (Original post by EEngWillow)
    I'm not actually sure about this, so hopefully someone will correct me if wrong.

    You can evaluate things like ln(-3), they just have a complex part:

    ln(-1) = i\pi (Easy to see from Euler's identity: e^{i\pi} +1 = 0, )

    so ln(-3) = ln(3) + ln(-1) = 3 + i\pi.

    Likewise, 2ln(-3) = 2ln(3) + 2ln(-1)= ln(9) + 2i\pi

    The real part of your answer is still as expected (ln(9)), but there's an added imaginary part.

    If you haven't come across complex numbers yet that might not make sense. If I'm honest I'm still not sure why we can't just say "well -3 squared is +9, let's ignore the complex part"!
    Woooaaaahhhhh hold on a minute!

    The logarithm of a negative (or complex) number isn't well-defined. For example, e^{3i\pi}+1=0 and e^{-2139i\pi}+1=0 and so on. So we could just as validly say that \ln(-1) = 3i\pi or \ln(-1) = -2139i\pi, but it can't be both, can it?

    We can "force it" to be well-defined, though. If z=re^{i\theta} with -\pi < \theta \le \pi we can say the "principal logarithm" is \ln r + i\theta.

    BUT this is a C3 question, and there's no way that complex numbers (FP1) or complex logarithms (degree-level) are going to come up in any questions... and as such, the only solution to the OP's problem is x=+6. If there were any more solutions, then we'd have some factor of i kicking around, which doesn't happen here.
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    (Original post by nuodai)
    Woooaaaahhhhh hold on a minute!

    The logarithm of a negative (or complex) number isn't well-defined. For example, e^{3i\pi}+1=0 and e^{-2139i\pi}+1=0 and so on. So we could just as validly say that \ln(-1) = 3i\pi or \ln(-1) = -2139i\pi, but it can't be both, can it?

    We can "force it" to be well-defined, though. If z=re^{i\theta} with -\pi < \theta \le \pi we can say the "principal logarithm" is \ln r + i\theta.

    BUT this is a C3 question, and there's no way that complex numbers (FP1) or complex logarithms (degree-level) are going to come up in any questions... and as such, the only solution to the OP's problem is x=+6. If there were any more solutions, then we'd have some factor of i kicking around, which doesn't happen here.
    Aha, thanks nuodai, that makes a lot more sense now. The 'forcing' thing comes from Engineering, we don't really do proper maths...

    I did forget to say explicitly that this was never going to be needed for a C3 question, although I thought I had! :s
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    (Original post by ViralRiver)
    x = +6
    which equals to 6
 
 
 
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