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Math Help - Logs and more logs....

  1. #1
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    Logs and more logs....

    I have to find an expression for the nth term for all these sequences and write the expressions in the form \frac {p}{q} where p, q, exists on integers.

    Here are the sequences:
    \log_2{8}, \log_4{8}, \log_8{8}, \log_{16}{8}, \log_{32}{8}, \log_{64}{8}, \log_{128}{8}

    \log_{m}{m^k}, \log_{m^2}{m^k}, \log_{m^3}{m^k}, \log_{m^4}{m^k}, \log_{m^5}{m^k}, \log_{m^6}{m^k}, \log_{m^7}{m^k}

    I don't know where to begin or how to approach this. Help?
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  2. #2
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    Quote Originally Posted by tinytiger View Post
    I have to find an expression for the nth term for all these sequences and write the expressions in the form \frac {p}{q} where p, q, exists on integers.

    Here are the sequences:
    \log_2{8}, \log_4{8}, \log_8{8}, \log_{16}{8}, \log_{32}{8}, \log_{64}{8}, \log_{128}{8}

    \log_{m}{m^k}, \log_{m^2}{m^k}, \log_{m^3}{m^k}, \log_{m^4}{m^k}, \log_{m^5}{m^k}, \log_{m^6}{m^k}, \log_{m^7}{m^k}

    I don't know where to begin or how to approach this. Help?
    I'll take the first example:

    1. You are supposed to know that 8 = 2^3

    2. Use the base-change-formula of logarithms:

    \log_2(8) = \dfrac{\ln(8)}{\ln(2)} = \dfrac{3 \ln(2)}{\ln(2)} = 3

    \log_4(8) = \dfrac{\ln(8)}{\ln(4)} = \dfrac{3 \ln(2)}{ 2\ln(2)} = \dfrac32

    \log_8(8) = \dfrac{\ln(8)}{\ln(8)} = \dfrac{3 \ln(2)}{ 3\ln(2)} = \dfrac33 = 1

    \log_{16}(8) = \dfrac{\ln(8)}{\ln(16)} = \dfrac{3 \ln(2)}{ 4\ln(2)} = \dfrac34

    Do you see the pattern?

    The second example should be done in just the same way.
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  3. #3
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    So for the first sequence the expression would be:

    \log_{2^n}(8) = \dfrac {\ln{2^3}}{\ln{2^n}} = \dfrac{3 \ln(2)}{n \ln(2)} = \dfrac {3}{n}


    And the last sequence the expression would be:

    \log_{m^n}{m^k} = \dfrac {\ln{m^k}}{\ln{m^n}} = \dfrac{k \ln(m)}{n \ln(m)} = \dfrac {k}{n}

    It makes sense now.
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  4. #4
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    Quote Originally Posted by tinytiger View Post
    So for the first sequence the expression would be:

    \log_{2^n}(8) = \dfrac {\ln{2^3}}{\ln{2^n}} = \dfrac{3 \ln(2)}{n \ln(2)} = \dfrac {3}{n}


    And the last sequence the expression would be:

    \log_{m^n}{m^k} = \dfrac {\ln{m^k}}{\ln{m^n}} = \dfrac{k \ln(m)}{n \ln(m)} = \dfrac {k}{n}

    It makes sense now.
    Correct!
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  5. #5
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    I just have one last question.

    I have to obtain the third answer from the first two answers.

    \log_4{64} and \log_8{64} must some how be calculated to get \log_{32}{64}.

    I know that \log_4{64} = 3 and \log_8{64} = 2 and \log_{32}{64} = 1.2. I also recognize that the bases in the first 2 logarithms, 4 which is 2^2, and 8, which is 2^3, must have a connection with the last one, 32, which is 2^5. I can't seem to figure it out though.
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  6. #6
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    Quote Originally Posted by tinytiger View Post
    I just have one last question.

    I have to obtain the third answer from the first two answers.

    \log_4{64} and \log_8{64} must some how be calculated to get \log_{32}{64}.

    I know that \log_4{64} = 3 and \log_8{64} = 2 and \log_{32}{64} = 1.2. I also recognize that the bases in the first 2 logarithms, 4 which is 2^2, and 8, which is 2^3, must have a connection with the last one, 32, which is 2^5. I can't seem to figure it out though.
    You probably have seen that \ln(4)+\ln(8)=2\ln(2)+3\ln(2)=5\ln(2)=\ln(32)

    Using the same method as in the previous posts you get:

    \log_{32}(64)=\dfrac{6\ln(2)}{5\ln(2)}=\dfrac{6\ln  (2)}{2\ln(2)+3\ln(2)}= \dfrac65
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  7. #7
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    I, personally, wouldn't have used "ln", or the change of base formula for thes problems.

    Instead use the basic definition of log_a(x) as the inverse to a^x, and vice-versa. That means that log_a(a^x)= x.

    So log_2(8)= log_2(2^3)= 3 immediately.

    And, in general, for any n, 2^3= (2^3)^{n(1/n)}= 2^{3n(1/n)}= (2^n)^{3/n} and so log_{2^n}(2^3)= log_{2^n}((2^n)^{3/n})= \frac{3}{n}.

    m^3= (m^k)^{n(1/n)}= m^{nk/n}= (m^n)^{k/n} so log_{m^n}(m^k)= log_{m^n}((m^n)^{k/n})= k/n.
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  8. #8
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    Thanks for the help!
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