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Math Help - Please guide me along this logarithms sum(on expressing my answers in terms of p & q)

  1. #1
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    Please guide me along this logarithms sum(on expressing my answers in terms of p & q)

    Hi ,
    I got stuck doing this log question:
    Given that log base 2 x=p and log base 4 y =q, express answer in terms of p and q.

    a) log base x 4y
    b)x^2y

    Help is appreciated because I need to prepare for an upcoming test. Thanks
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  2. #2
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    Quote Originally Posted by AeroScizor View Post
    Hi ,
    I got stuck doing this log question:
    Given that log base 2 x=p and log base 4 y =q, express answer in terms of p and q.

    a) log base x 4y
    b)x^2y

    Help is appreciated because I need to prepare for an upcoming test. Thanks
    Hi AeroScizor,

    Part (a)

    log_2x=p

    Now slide the "2" over to the "p" and drop the log

    x=_2p

    x=2^p

    log_4y=q\ \Rightarrow\ y=4^q\ \Rightarrow\ 4y=(4)4^q=4^{q+1}


    The change of base formula is log_ax=\frac{log_bx}{log_ba}

    log_x(4y)=\frac{log_2(4y)}{log_2(x)}

    log_2(4y)=log_2\left(4^{q+1}\right)=z\ \Rightarrow\ 2^z=4^{q+1}=2^{2(q+1)}

    Therefore log_2(4y)=z=2(q+1)

    Hence log_x(4y)=\frac{2(q+1)}{p}

    Part (b)

    x^2y=\left(2^p\right)2^p\left(4^q\right)=2^{p+p}\l  eft(2^2\right)^q=2^{2p}2^{2q}=2^{2p+2q}=2^{2(p+q)}  =4^{p+q}


    If you meant x^{2y}

    2y=(2)4^q=(2)2^{2q}=2^{2q+1}

    x^{2y}=\left(2^p\right)^{2^{2q+1}}

    Edit, sorry! I wrote the logs inverted earlier!!
    Last edited by Archie Meade; February 25th 2010 at 07:09 AM. Reason: error
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  3. #3
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    Hello, AeroScizor!

    I got a different answer for (a).


    Given: . \log_2(x)\:=\:p\:\text{ and }\:\log_4(y) \:=\:q

    (a) Express \log_x(4y) in terms of p\text{ and }q.
    Note that: . \log_4(y) \:=\:q \quad\Rightarrow\quad 4^q \:=\:y \quad\Rightarrow\quad (2^2)^q \:=\:y \quad\Rightarrow\quad 2^{2q} \:=\:y

    . . Hence: . \log_2(y) \:=\:2q


    Base-change Formula: . \log_x(4y) \;=\;\frac{\log_2(4y)}{\log_2(x)} \quad\Rightarrow\quad \frac{\log_2(4) + \log_2(y)}{\log_2(x)} \;=\;\frac{2 + 2q}{p}

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, AeroScizor!

    I got a different answer for (a).

    Note that: . \log_4(y) \:=\:q \quad\Rightarrow\quad 4^q \:=\:y \quad\Rightarrow\quad (2^2)^q \:=\:y \quad\Rightarrow\quad 2^{2q} \:=\:y

    . . Hence: . \log_2(y) \:=\:2q


    Base-change Formula: . \log_x(4y) \;=\;\frac{\log_2(4y)}{\log_2(x)} \quad\Rightarrow\quad \frac{\log_2(4) + \log_2(y)}{\log_2(x)} \;=\;\frac{2 + 2q}{p}
    Thanks for your reply, I finally get the concept now! Cheers.
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    Hi AeroScizor,

    Part (a)

    log_2x=p

    Now slide the "2" over to the "p" and drop the log

    x=_2p

    x=2^p

    log_4y=q\ \Rightarrow\ y=4^q\ \Rightarrow\ 4y=(4)4^q=4^{q+1}


    The change of base formula is log_ax=\frac{log_bx}{log_ba}

    log_x(4y)=\frac{log_2(4y)}{log_2(x)}

    log_2(4y)=log_2\left(4^{q+1}\right)=z\ \Rightarrow\ 2^z=4^{q+1}=2^{2(q+1)}

    Therefore log_2(4y)=z=2(q+1)

    Hence log_x(4y)=\frac{2(q+1)}{p}

    Part (b)

    x^2y=\left(2^p\right)2^p\left(4^q\right)=2^{p+p}\l  eft(2^2\right)^q=2^{2p}2^{2q}=2^{2p+2q}=2^{2(p+q)}  =4^{p+q}


    If you meant x^{2y}

    2y=(2)4^q=(2)2^{2q}=2^{2q+1}

    x^{2y}=\left(2^p\right)^{2^{2q+1}}

    Edit, sorry! I wrote the logs inverted earlier!!
    Thanks for your reply, I get the concept now. Cheers
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