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Math Help - Assignment natural logs

  1. #1
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    Assignment natural logs

    I've gotten most of my questions answered by looking at the answers people have posted for people in my class, but I'm still confused with two questions. Any help would be very much appreciated!

    1. Express the function y= 5^x as an equivalent function with base 3.

    2. Prove logab = logcb / logca.

    And we were given a hint for this question. Let x= logab, y= logcb and z= logca. Convert these to exponential for...

    The work I've done so far for #2 is:
    a^x=b
    c^y=b
    c^z=a

    but that's about all I have and I'm not sure where exactly to go from there in the question.


    Thank-you for any help!
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  2. #2
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    Quote Originally Posted by jmailloux View Post
    I've gotten most of my questions answered by looking at the answers people have posted for people in my class, but I'm still confused with two questions. Any help would be very much appreciated!

    1. Express the function y= 5^x as an equivalent function with base 3.
    This has been answered in this thread.

    RonL
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  3. #3
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    Quote Originally Posted by jmailloux View Post

    2. Prove logab = logcb / logca.
    Suppose a,b,c>0

    let b=a^y, then:

    \log_c(b)=y\log_c(a),

    but y=log_a(b), so

    \log_c(b)=\log_a(b)\log_c(a)

    or:

    \log_a(b)=\log_c(b)/\log_c(a)

    RonL
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  4. #4
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    still confused

    thanks for the help. I understand #2, but I'm still having problems understanding how to arrive at the answer for #1. If someone could please explain he steps to get to the answer, it would be much appreciated.

    what i've done for #1 is
    3^z= 5^x
    log3^z= log5^x
    zlog3= xlog5
    z= xlog5/log3

    therefore, y=5^x as an equivalent with base 3 would be

    y= 3^xlog5/log3

    is this right??? its slightly different from the answers on the other posts in the calculus forum.. but i don't understand how to arrive at those answers. Any help would be SOO great b/c i have such a headache from trying to finish this assignment. THANK YOU
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  5. #5
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    Quote Originally Posted by jmailloux View Post
    thanks for the help. I understand #2, but I'm still having problems understanding how to arrive at the answer for #1. If someone could please explain he steps to get to the answer, it would be much appreciated.

    what i've done for #1 is
    3^z= 5^x
    log3^z= log5^x
    zlog3= xlog5
    z= xlog5/log3

    therefore, y=5^x as an equivalent with base 3 would be

    y= 3^xlog5/log3
    1. Put more brackets in to make what you mean clearer.

    2. if you mean:

    y=3^{x (\log(5)/\log(3))},

    then as \log_3(5)=\log(5)/\log(3) (by part 2 of your question) this last equation is equivalent to:

    y=3^{x \log_3(5)},

    which is what the other answer gave, so yes this would be correct, though
    clumsy.

    RonL
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  6. #6
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    still confused

    I don't understand this part of the answer

    log3(5)= log(5)/log(3)

    How is log3(5) equal to that? It's probably the simplest thing to understand, but I don't get the process there. If someone could explain that to me, that would be great. Or if you have another "NON-CLUMSY" way of answering the question that you think I would understand, that would be great!!!
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  7. #7
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    Quote Originally Posted by jmailloux View Post
    I don't understand this part of the answer

    log3(5)= log(5)/log(3)

    How is log3(5) equal to that?
    It is the change of base for logarithms formula. You will have
    seen it in this question from your original post:

    "2. Prove logab = logcb / logca."

    in this case c is whatever the base for log in your expression

    "log(5)/log(3)"

    is, presumably e or 10 (the two most common bases).

    It's probably the simplest thing to understand, but I don't get the process there. If someone could explain that to me, that would be great. Or if you have another "NON-CLUMSY" way of answering the question that you think I would understand, that would be great!!!
    I have given a less clumsy way already, and if I recall correctly it used
    the result that:

    A=B^{\log_B(A)}

    which is close to the definition of what \log_B(A) means.

    RonL
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