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Math Help - logarithm problems

  1. #1
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    logarithm problems

    evaluate
    1.1.5*4^(log (base 4*sqrt(2))(3-sqrt(6))-6log(base 8)(sqrt(3)-sqrt(2))

    2.5^(log(base(1/5)(1/2))+log(base sqrt(2))(4/(sqrt(7)+sqrt(3))+log(base 1/2)(1/(10+2sqrt(21))

    please help me in doing these problems i dont know latex
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  2. #2
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    Hello, prasum!

    It would help if you learned to use the Space Bar.
    I must guess what you meant . . .



    \text{2. Evaluate: }\;5^{\log_{\frac{1}{5}}\left(\frac{1}{2}\right)}+ \log_{\sqrt{2}}\!\left(\dfrac{4}{\sqrt{7}-\sqrt{3}}\right) + \log_{\frac{1}{2}}\!\left(\dfrac{1}{10+2\sqrt{21}}  \right)

    Each of the three terms takes an enormous amount of simplifying!


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    The exponent of the first term: . \log_{\frac{1}{5}}\!\!\left(\frac{1}{2}\right) \,=\,P

    . . \left(\frac{1}{5}\right)^P \:=\:\frac{1}{2}\quad\Rightarrow\quad \left(5^{-1}\right)^P \:=\:2^{-1} \quad\Rightarrow\quad 5^{-P} \:=\:2^{-1}


    Take logs (base 5): . \log_5\!\left(5^{\text{-}P}\right) \:=\:\log_5\!\left(2^{\text{-}1}\right) \quad\Rightarrow\quad \text{-}P\log_5(5) \:=\:\text{-}\log_5(2)

    . . Hence: . P \:=\:\log_5(2) \quad\Rightarrow\quad \log_{\frac{1}{5}}\left(\frac{1}{2}\right) \:=\:\log_5(2)


    The first term is: . 5^{\log_5(2)} \;=\;2


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    In the second term, rationalize the denominator.

    . . \dfrac{4}{\sqrt{7}-\sqrt{3}} \cdot \dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}} \;=\;\dfrac{4(\sqrt{7} + \sqrt{3})}{7-3} \:=\:\sqrt{7} + \sqrt{3}


    We have: . \log_{\sqrt{2}}(\sqrt{7}+\sqrt{3}) \:=\:Q<br />
\quad\Rightarrow\quad (\sqrt{2})^Q \:=\:\sqrt{7}+\sqrt{3}

    . . (2^{\frac{1}{2}})^Q \:=\:\sqrt{7} + \sqrt{3} \quad\Rightarrow\quad 2^{\frac{Q}{2}} \:=\:\sqrt{7} + \sqrt{3}


    Take logs (base 2): . \log_2\!\left(2^{\frac{Q}{2}}\right) \:=\:\log_2(\sqrt{7}+\sqrt{3})

    . . \frac{Q}{2}\log_2(2) \:=\:\log_2(\sqrt{7}+\sqrt{3}) \quad\Rightarrow\quad Q \:=\:2\log_2(\sqrt{7}+\sqrt{3})

    . . Q \:=\:\log_2(\sqrt{7} + \sqrt{3})^2


    The second term is: . 2\log_2(\sqrt{7} + \sqrt{3})


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    The thrd term is: . \log_{\frac{1}{2}}\!\left(\dfrac{1}{10+2\sqrt{21}}  \right) \:=\:\log_{\frac{1}{2}}\!\dfrac{1}{(\sqrt{7}+\sqrt  {3})^2}


    We have: . \log_{\frac{1}{2}}(\sqrt{7}+\sqrt{3})^{-2} \:=\: R \quad\Rightarrow\quad \left(\frac{1}{2}\right)^R \:=\:(\sqrt{7}+\sqrt{3})^{-2}

    . . \left(2^{-1}\right)^R \:=\:(\sqrt{7}+\sqrt{3})^{-2} \quad\Rightarrow\quad 2^{-R} \:=\:(\sqrt{7}+\sqrt{3})^{-2}


    Take logs (base 2): . \log_2(2^{-R}) \:=\:\log_2(\sqrt{7}+\sqrt{3})^{-2}

    . . -R\log_2(2) \:=\:-2\log_2(\sqrt{7}+\sqrt{3})

    . . . . . . . . R \;=\;\2\log_2(\sqrt{7}+\sqrt{3})


    The third term is: . 2\log_2(\sqrt{7}+\sqrt{3})


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    The original expression is: . 2 + 2\log_2\!\left(\sqrt{7}+\sqrt{3}\right) + 2\log_2\!\left(\sqrt{7}+\sqrt{3}\right)

    . . =\;\log_2(4) + 4\log_2\!\left(\sqrt{7}+\sqrt{3}\right)

    . . =\;\log_2\!\bigg[4\left(\sqrt{7}+\sqrt{3}\right)^4\bigg]

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    But check my work . . . please!
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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, prasum!

    It would help if you learned to use the Space Bar.
    I must guess what you meant . . .




    Each of the three terms takes an enormous amount of simplifying!


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    The exponent of the first term: . \log_{\frac{1}{5}}\!\!\left(\frac{1}{2}\right) \,=\,P

    . . \left(\frac{1}{5}\right)^P \:=\:\frac{1}{2}\quad\Rightarrow\quad \left(5^{-1}\right)^P \:=\:2^{-1} \quad\Rightarrow\quad 5^{-P} \:=\:2^{-1}


    Take logs (base 5): . \log_5\!\left(5^{\text{-}P}\right) \:=\:\log_5\!\left(2^{\text{-}1}\right) \quad\Rightarrow\quad \text{-}P\log_5(5) \:=\:\text{-}\log_5(2)

    . . Hence: . P \:=\:\log_5(2) \quad\Rightarrow\quad \log_{\frac{1}{5}}\left(\frac{1}{2}\right) \:=\:\log_5(2)


    The first term is: . 5^{\log_5(2)} \;=\;2


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    In the second term, rationalize the denominator.

    . . \dfrac{4}{\sqrt{7}-\sqrt{3}} \cdot \dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}} \;=\;\dfrac{4(\sqrt{7} + \sqrt{3})}{7-3} \:=\:\sqrt{7} + \sqrt{3}


    We have: . \log_{\sqrt{2}}(\sqrt{7}+\sqrt{3}) \:=\:Q<br />
\quad\Rightarrow\quad (\sqrt{2})^Q \:=\:\sqrt{7}+\sqrt{3}

    . . (2^{\frac{1}{2}})^Q \:=\:\sqrt{7} + \sqrt{3} \quad\Rightarrow\quad 2^{\frac{Q}{2}} \:=\:\sqrt{7} + \sqrt{3}


    Take logs (base 2): . \log_2\!\left(2^{\frac{Q}{2}}\right) \:=\:\log_2(\sqrt{7}+\sqrt{3})

    . . \frac{Q}{2}\log_2(2) \:=\:\log_2(\sqrt{7}+\sqrt{3}) \quad\Rightarrow\quad Q \:=\:2\log_2(\sqrt{7}+\sqrt{3})

    . . Q \:=\:\log_2(\sqrt{7} + \sqrt{3})^2


    The second term is: . 2\log_2(\sqrt{7} + \sqrt{3})


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    The thrd term is: . \log_{\frac{1}{2}}\!\left(\dfrac{1}{10+2\sqrt{21}}  \right) \:=\:\log_{\frac{1}{2}}\!\dfrac{1}{(\sqrt{7}+\sqrt  {3})^2}


    We have: . \sqrt{7}+\sqrt{3})^{-2} " alt="\log_{\frac{1}{2}}(\sqrt{7}+\sqrt{3})^{-2} \:=\: R \quad\Rightarrow\quad \left(\frac{1}{2}\right)^R \:=\\sqrt{7}+\sqrt{3})^{-2} " />

    . . \sqrt{7}+\sqrt{3})^{-2} \quad\Rightarrow\quad 2^{-R} \:=\\sqrt{7}+\sqrt{3})^{-2} " alt="\left(2^{-1}\right)^R \:=\\sqrt{7}+\sqrt{3})^{-2} \quad\Rightarrow\quad 2^{-R} \:=\\sqrt{7}+\sqrt{3})^{-2} " />


    Take logs (base 2): . \log_2(2^{-R}) \:=\:\log_2(\sqrt{7}+\sqrt{3})^{-2}

    . . -R\log_2(2) \:=\:-2\log_2(\sqrt{7}+\sqrt{3})

    . . . . . . . . R \;=\;\2\log_2(\sqrt{7}+\sqrt{3})


    The third term is: . 2\log_2(\sqrt{7}+\sqrt{3})


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    The original expression is: . 2 + 2\log_2\!\left(\sqrt{7}+\sqrt{3}\right) + 2\log_2\!\left(\sqrt{7}+\sqrt{3}\right)

    . . =\;\log_2(4) + 4\log_2\!\left(\sqrt{7}+\sqrt{3}\right)

    . . =\;\log_2\!\bigg[4\left(\sqrt{7}+\sqrt{3}\right)^4\bigg]

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    But check my work . . . please!
    thanks
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