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Thread: log problem

  1. #1
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    log problem

    Given that $\displaystyle \log_4T=2+\log_2v,$express t in term of V.
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  2. #2
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    Quote Originally Posted by mastermin346 View Post
    Given that $\displaystyle \log_4T=2+\log_2v,$express t in term of V.

    hmmmm.....

    $\displaystyle \log_4T=2+\log_2v,$

    $\displaystyle T=4^{2+\log_2v}$
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  3. #3
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    Hello, pickslides!

    I just noticed that your answer can be simplified beyond all recognition . . .


    $\displaystyle T\:=\:4^{2+\log_2v}$

    $\displaystyle T \;=\;4^2\cdot 4^{\log_2v} \;=\;16\cdot(2^2)^{\log_2v} \;=\;16\cdot2^{2\log_2v} \;=\;16\cdot2^{\log_2(v^2)} \;=\;16v^2$



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  4. #4
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    Hello,

    Here is another way :

    We know that $\displaystyle \log_a b=\frac{\ln b}{\ln a}$
    From here, it's easy to see that $\displaystyle \log_2 v=2 \log_4(v^2)$

    And we know that $\displaystyle 2=\log_4 4^2=\log_4 16$, so that $\displaystyle 2+\log_2 v=\log_4 (16v^2)$

    ---> $\displaystyle \log_4 T=\log_4 (16v^2)\Rightarrow T=16v^2$
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