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Math Help - Write as single logarithm

  1. #1
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    Write as single logarithm

    Write (equation) as a single logarithm.

    Attached is the equation and the solution.

    Did I solve it correctly?
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  2. #2
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    no you did not solve it correctly, remember the rules of log,

    4\ln(x-3) is the same as \ln(x-3)^4

    and 2\ln(x) is the same as \ln(x)^2

    also
    \ln(x-3)^4 + \ln(x)^2 is the same as
    \ln(x-3)^4x^2
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    Ah. Is this correct?
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    and there it is, as a single logarithm, u can try to put any number back into the equation to check your working e.g. let x = 5

    if u put this 4\ln(x-3) + 2\ln(x) with x equals 5 into your calculator you will get 5.999 or something

    and if you put x=5 into the single logarithm \ln(x-3)^4x^2
    you will get the same 5.99 so you know its correct

    hope you understand
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    Thanks!
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  6. #6
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    Quote Originally Posted by mwok View Post
    Ah. Is this correct?
    There's just a problem with the x^4-3^4 part

    (x-3)^4=[(x-3)^2]^2=(x^2-6x+9)^2= (x^4+36x^2+81-12x^3-108x+18x^2)=(x^4-12x^3+54x^2-108x+81)

    But I think you can just keep it with (x-3)^4
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    Wow.

    Okay, so the final solution is attached.

    EDIT:
    Why does the "x" come out of ln(x)^2??
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  8. #8
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    because of the two infront of it, note that this only works with logs/ln
    2\ln(x) = \ln(x)^2
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    Quote Originally Posted by iLikeMaths View Post
    because of the two infront of it, note that this only works with logs/ln
    2\ln(x) = \ln(x)^2

    ....what happens to the "ln"?
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  10. #10
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    Quote Originally Posted by mwok View Post
    Wow.

    Okay, so the final solution is attached.

    EDIT:
    Why does the "x" come out of ln(x)^2??
    No, it's \ln(x^2)

    It's not really 4 \log(x-3) !=\log(x^4-3^4). I mean... stated this way, it looks like you didn't understand your mistake.

    What you wrote is :
    4 \log(x-3)=\log((x-3)^4)

    But (x-3)^4 !=x^4-3^4, this was your mistake.


    The final answer, \ln((x-3)^4x^2) is correct.
    But put as many parentheses as possible, because it's very easy to get wrong.
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  11. #11
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    Quote Originally Posted by iLikeMaths View Post
    because of the two infront of it, note that this only works with logs/ln
    2\ln(x) = \ln(x)^2
    It's not a very correct writing.
    Because one would understand it as "the square of the logarithm" whereas it is "the logarithm of the square"

    So it should be written \ln(x^2) (or \ln x^2, but it can still be ambiguous)
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    Ah I see how the exponents work now BUT why does x^2 come out of the ln(x^2)? What happens to the ln?
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  13. #13
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    Quote Originally Posted by mwok View Post
    Ah I see how the exponents work now BUT why does x^2 come out of the ln(x^2)? What happens to the ln?
    We have :
    \ln((x-3)^4)+\ln(x^2)

    Now use the rule \ln(a)+\ln(b)=\ln(ab) :

    =\ln[(x-3)^4 x^2]
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    Thanks!
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  15. #15
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    some times != means does not equal, well thats what i thought he/she meant
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