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Thread: 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 $\displaystyle \frac {p}{q}$ where $\displaystyle p$, $\displaystyle q$, exists on integers.

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

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

    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 $\displaystyle \frac {p}{q}$ where $\displaystyle p$, $\displaystyle q$, exists on integers.

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

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

    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 $\displaystyle 8 = 2^3$

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

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

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

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

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

    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:

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


    And the last sequence the expression would be:

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

    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:

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


    And the last sequence the expression would be:

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

    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.

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

    I know that $\displaystyle \log_4{64} = 3$ and $\displaystyle \log_8{64} = 2$ and $\displaystyle \log_{32}{64} = 1.2$. I also recognize that the bases in the first 2 logarithms, 4 which is $\displaystyle 2^2$, and 8, which is $\displaystyle 2^3$, must have a connection with the last one, 32, which is $\displaystyle 2^5$. 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.

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

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

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

    $\displaystyle \log_{32}(64)=\dfrac{6\ln(2)}{5\ln(2)}=\dfrac{6\ln (2)}{2\ln(2)+3\ln(2)}= \dfrac65$
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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 $\displaystyle log_a(x)$ as the inverse to $\displaystyle a^x$, and vice-versa. That means that $\displaystyle log_a(a^x)= x$.

    So $\displaystyle log_2(8)= log_2(2^3)= 3$ immediately.

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

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