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    ∫ 1/(4 -3(x)) dx = (-1/3) ln|4 -3(x)| + k => (-1/3) lnA | 4-3(x)| = (1/3) ln A/ |4-3(x)|

    ok im having 3 doubts about this example in my textbook

    1) where does the A come from?
    i was guessing since K is a constant , so K could have been ln(A) which is also a constant and as such, by log law; ln a + ln b = ln ab

    2)modulus is because of ln cant be negative?

    3) how and why did (-1/3) lnA | 4-3(x)| became 1/3) ln A/ |4-3(x)| ??
    this one i can't figure it out, can anyone explain this please?
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    (Original post by Carlos Nim)
    ∫ 1/(4 -3(x)) dx = (-1/3) ln|4 -3(x)| + k => (-1/3) lnA | 4-3(x)| = (1/3) ln A/ |4-3(x)|

    ok im having 3 doubts about this example in my textbook

    1) where does the A come from?
    i was guessing since K is a constant , so K could have been ln(A) which is also a constant and as such, by log law; ln a + ln b = ln ab
    Yep.

    2)modulus is because of ln cant be negative?
    Yep.

    3) how and why did (-1/3) lnA | 4-3(x)| became 1/3) ln A/ |4-3(x)| ??
    this one i can't figure it out, can anyone explain this please?
    We essentially have -\ln ax = \ln \frac{1}{ax} but if a is some constant that we don't particularly care too much about, it doesn't matter whether it is on the numerator or denominator as long as it's there hence we can just \frac{1}{a} \mapsto a without any effect and we get that -\ln ax = \ln \frac{a}{x}
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    wait a min, you say without any effect, it can't be true..... for e.g if a is 2 and x is 4
    1) - ln (2.4) = -2.079
    2) ln (1/(2.4)) = -2.079
    3) ln (2/4) = -0.6931
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    can you clarify what i edited above please
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    (Original post by Carlos Nim)
    can you clarify what i edited above please
    It's just an arbitrary constant, we usually dont need to care about what it is so we can manipulate it like that.
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    (Original post by RDKGames)
    It's just an arbitrary constant, we usually dont need to care about what it is so we can manipulate it like that.
    so then i shouldn't be penalise in the exams if i write (1/3) ln [1/(A|4-3(x)|)] instead of (1/3) ln [A/ |4-3(x)|]
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    (Original post by Carlos Nim)
    so then i shouldn't be penalise in the exams if i write (1/3) ln [1/(A|4-3(x)|)] instead of (1/3) ln [A/ |4-3(x)|]
    It's perfectly fine in this context, just the point is that it's neater to have A in the numerator than the denominator if possible.
 
 
 
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