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    Understand everything except I didn't get lnA for part b as it says in the markscheme (second line). Isn't this just supposed to be a constant? Don't see why it's lnA and not just A.

    Thanks :-)
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    (Original post by Chickenslayer69)
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    Understand everything except I didn't get lnA for part b as it says in the markscheme (second line). Isn't this just supposed to be a constant? Don't see why it's lnA and not just A.

    Thanks :-)
    It's just neater this way. Note that you can let your constant be c but then how would you combine \ln(f(x))+c under a single logarithm?? Now \ln(A) is still an arbitrary constant if A>0, and the natural logarithm function means it can be any value you want it to be so there's no problem with us using it. Now since we can use it, we can combine the logarithms which makes the working out easier.
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    (Original post by RDKGames)
    It's just neater this way. Note that you can let your constant be c but then how would you combine \ln(f(x))+c under a single logarithm?? Now \ln(A) is still an arbitrary constant if A>0, and the natural logarithm function means it can be any value you want it to be so there's no problem with us using it. Now since we can use it, we can combine the logarithms which makes the working out easier.
    Oh, so it's not that it has to be that, it just can be? I suspected that but didn't know for sure. Thanks for clarifying :-)
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    (Original post by Chickenslayer69)
    Oh, so it's not that it has to be that, it just can be? I suspected that but didn't know for sure. Thanks for clarifying :-)
    No it doesn't have to be. In fact for \ln[f(y)]=\ln[f(x)]+c you simply get:

    \displaystyle f(y)=e^{\ln[f(x)]+c}=e^{\ln[f(x)]} \cdot e^{c} = e^c \cdot f(x) now you can just let A=e^c since e^c is just some constant number anyway, and achieve the same answer as in the book.

    The way they did it is not a requirement, just a preference, I suppose.
 
 
 
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