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Math Help - Logarithmic Functions

  1. #1
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    Logarithmic Functions

    This problem actually does not involve calculus, but eventually we will need to take derivatives of this expressions. The instructions indicate: Use logarithmic rules(like product,power,quotient rules) to rewrite each expression in terms of log3 2 and log3 5. Here is the problem: log3 270. (The 3 is lowercase, 270 is larger). The answer is: 3+log3 2+log3 5. If someone can illustrate how to get this, I would appreciate it, Thanks for the help.
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    Quote Originally Posted by kdogg121 View Post
    This problem actually does not involve calculus, but eventually we will need to take derivatives of this expressions. The instructions indicate: Use logarithmic rules(like product,power,quotient rules) to rewrite each expression in terms of log3 2 and log3 5. Here is the problem: log3 270. (The 3 is lowercase, 270 is larger). The answer is: 3+log3 2+log3 5. If someone can illustrate how to get this, I would appreciate it, Thanks for the help.
    log_3(270) = log_3(2 \cdot 3^3 \cdot 5)

    = log_3(2) + log_3(3^3) + log_3(5)

    = log_3(2) + 3 \cdot log_3(3) + log_3(5)

    = log_3(2) + 3 \cdot 1 + log_3(5)

    = 3 + log_3(2) + log_3(5)

    -Dan
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  3. #3
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    What about this one: We have to solve the given equation for x.

    3= 2+5e^-4x I took the natural log (ln) for all the terms so that the e term would cancel out and I was left with ln3= ln2-4X I then isolated the x term and had ln1= -4x and got an answer of x= ln1/-4 which equals 0. The book has the answer as: ln5/4 =0.402 approx. What did I do wrong?
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  4. #4
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    Hello,

    \log(a-b) \; \text{is not equal to} \; \log(a)-\log(b)

    Firstly, simplify 2 and 3.
    Then you can put the logarithm, and use the following properties to isolate x :

    \log(ab)=\log(a)+log(b)

    \log(e^a)=a
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  5. #5
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    I'm sorry I still don't understand how that will work out, could someone illustrate what should be done? My professor said to cancel the e by using the ln on each term. Could you show how the problem is worked out?
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  6. #6
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    3= 2+5e^-4x

    This is the same as 1=5e^(-4x)

    You can't compose a sum with the logarithm, it's a nonsens and unuseful, because you won't be able to isolate x.

    So now you can compose with the logarithm.

    log(1)=log(5 e^{-4x})

    As mentioned above (i gave you the general formulae you needed...) :

    log(5 e^{-4x})=log(5)+log(e^{-4x})=log(5)-4x

    => the equation will now be :

    0=log(5)-4x

    And you have x ;-)
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  7. #7
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    Ok, I follow what you are doing up to the end, but I think I need to have X isolated, to solve for it. What I did was subtract Log5 from the left side of the equation, which would give you -Log5= -4X I then divided both sides by -4 and the answer I got when computing this on my calculator was different than the book's answer of: .402 How do I get this?
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  8. #8
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    Hello,

    You have to differentiate ln and log (used on calculators for ln in basis 10). Otherwise, if you type ln(5)/4, you will have the correct answer :-)
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  9. #9
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    I was wondering whether I have the correct answer for this problem:

    ln of cubed root of x^2-x

    I have as an answer: 1/3(2lnx+lnx) Is this correct? Thanks...
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