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Math Help - Logarithms

  1. #1
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    Logarithms

    Given that x=log_2a and y=log_2b, express the following in terms of y and x

    (ii) log_{2a}\sqrt{ab^3}

    My attempt:

    x=log_2a
    a=2^x

    y=log_2b
    b=2^y

    log_{2a}\sqrt{ab^3}= log_{2(2^x)}\sqrt{2^x(2^{3y})}

    = log_{2^{2x+1}}\sqrt{2^x}+log_{2^{2x+1}}\sqrt{2^{3y  }}
    Last edited by Punch; April 17th 2010 at 09:02 PM.
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  2. #2
    Super Member Bacterius's Avatar
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    Why do you make it so complicated ?

    x = \log_2{(a)} \Rightarrow a = 2^x
    y = \log_2{(b)} \Rightarrow b = 2^y

    \log_2{\left ( a \sqrt{ab^3} \right )}

    = \log_2{\left (2^x \sqrt{2^x (2^y)^3} \right )}

    = \log_2{\left (2^x \sqrt{2^x 2^{3y}} \right )}

    = \log_2{\left (2^x \sqrt{2^{x + 3y}} \right )}

    = \log_2{\left (2^x \times 2^{\frac{x + 3y}{2}} \right )}

    = \log_2{\left (2^{x + \frac{x + 3y}{2}} \right )}

    = x + \frac{x + 3y}{2}

    = \frac{2x}{2} + \frac{x + 3y}{2}

    = \frac{2x + x + 3y}{2}

    = \frac{3x + 3y}{2}

    = \frac{3(x + y)}{2}

    = \frac{3}{2}(x + y)
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  3. #3
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    x=log_2a
    a=2^x

    y=log_2b
    b=2^y
    It is excellent that you have memorized this transformation. It seems, however, that you have not learned how to use the information in the section.

    Quote Originally Posted by Punch View Post
    Given that x=log_2a and y=log_2b, express the following in terms of y and x

    (ii) log_2a\sqrt{ab^3}

    My attempt:

    log_2a\sqrt{ab^3}= log_{2(2^x)}\sqrt{2^x(2^{3y})}


    1) You should not have done ANYTHING with the base of the logartihm. All the logarithms are Base 2.

    2) There is no such thing as a superscript on a logarithm. Perhaps it was just a codign error.
    = log_2^{2x+1}\sqrt{2^x}+log_2^{2x+1}\sqrt{2^{3y}}
    3) How about if we just use rules of logarithms to solve this.

    log_{2}(a\sqrt{ab^3})

    log_{2}((a)\cdot(\sqrt{ab^3}))

    log_{2}(a) + log_{2}(\sqrt{ab^3})

    log_{2}(a) + \frac{1}{2}log_{2}(ab^3)

    log_{2}(a) + \frac{1}{2}(log_{2}(a) + log_{2}(b^3))

    log_{2}(a) + \frac{1}{2}(log_{2}(a) + 3\cdot log_{2}(b))

    log_{2}(a) + \frac{1}{2} \cdot log_{2}(a) + \frac{3}{2}\cdot log_{2}(b)

    You can finish?
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  4. #4
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    You need not convert in to index form. Using the laws of logarithm you can simplify.





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  5. #5
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    Sorry but there was a typo error on the question i typed...

    it should be

    Given that x=log_2a and y=log_2b, express the following in terms of y and x

    (ii) log_{2a}\sqrt{ab^3}
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  6. #6
    Super Member Bacterius's Avatar
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    \log_{2a}{\left ( \sqrt{ab^3} \right )}

    = \log_{2^{x + 1}}{\left ( 2^{\frac{x + 3y}{2}} \right )}

    = \log_{2^{x + 1}}{\left ( 2^{\frac{x + 3y}{2}} \right )}

    = \frac{x + 3y}{2} \times \log_{2^{x + 1}}{\left ( 2 \right )}

    = \frac{x + 3y}{2} \times \frac{\log_2{(2)}}{\log_2 {\left ( 2^{x + 1}\right ) }}

    = \frac{x + 3y}{2} \times \frac{1}{x + 1}

    = \frac{x + 3y}{2(x + 1)}

    = \frac{1}{2} \times \frac{x + 3y}{x + 1}

    Does it make sense ?
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  7. #7
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    Hello, Punch!

    Given that: . \begin{Bmatrix} <br />
x&=& \log_2a  \\ y &=& \log_2b \\ \end{Bmatrix} .[1]

    express the following in terms of x\text{ and }y:

    . .  (ii)\;\;\log_{2a}\sqrt{ab^3}

    Let: . \log_{2a}\sqrt{ab^3} \:=\:P .[2]

    Then: . (2a)^P \;=\;\left(ab^3\right)^{\frac{1}{2}}


    Take logs (base 2): . \log_2\left(2a\right)^P \;=\;\log_2\left(ab^3\right)^{\frac{1}{2}}

    . . . . . . . . . . . . . . P\cdot\log_2(2a) \;=\;\frac{1}{2}\log_2\left(ab^3\right)

    . . . . . . . . P\cdot\bigg[\log_2(2) + \log_2a\bigg] \;=\;\frac{1}{2}\bigg[\log_2a + \log_2\left(b^3\right)\bigg]

    . . . . . . . . . . . P\cdot\bigg[1 + \log_2a\bigg] \;=\;\frac{1}{2}\bigg[\log_2a + 3\log_2b\bigg]

    Multiply by 2: . P\cdot2\bigg[1 + \log_2a\bigg] \;=\;\log_2a + 3\log_2b

    . . . . . . . . . . . . . . . . . . . P \;=\;\frac{\log_2a + 3\log_2b}{2(1+\log_2a)} .[3]


    Equate [2] and [3]: . \log_{2a}\sqrt{ab^3} \;=\;\frac{\log_2a + 3\log_2b}{2(1 + \log_2a)}

    . . . Substitute [1]: . \log_{2a}\sqrt{ab^3} \;=\;\frac{x+3y}{2(1+x)}

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  8. #8
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    Quote Originally Posted by Punch View Post
    Sorry but there was a typo error on the question i typed...

    it should be

    Given that x=log_2a and y=log_2b, express the following in terms of y and x

    (ii) log_{2a}\sqrt{ab^3}
    Using the change of base, rewrite the above problem.
    [1/2log(ab^3) to the basr 2]/[log(2a) to the base2]
    Now use the laws of logarithm to simplify.
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