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Math Help - Logarithm Simplification Help. Any aid appreciated.

  1. #1
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    Logarithm Simplification Help. Any aid appreciated.

    Hi everyone,

    I have been working with introductory logarithms recently, and I came across these two problems that I just could not solve. They are:

    The questions are:

    Simplify

    Log3 (4 to the power of 2) + 3 log3 2

    4 log3 2 2 log3 6 + 2

    Any help simplifying these equations will be very much appreciated!

    Thanks,

    Nathaniel
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  2. #2
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    \log_3{(4^2)} + 3\log_3{2} = \log_3{(2^4)} + 3\log_3{2}

     = 4\log_3{2} + 3\log_3{2}

     = 7\log_3{2}.
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  3. #3
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    Hello, BinaryBoy!

    \text{Simplify: }\;4\log_3(2) - 2\log_3(6) + 2


    4\log_3(2) - 2\log_3(6) + 2 \;=\;\log_3\left(2^4\right) - \log_3\left(6^2\right) + 2

    . . . . . . . . . . . . . . . . . . =\;\log_3(16)-\log_3(36) + 2

    . . . . . . . . . . . . . . . . . . =\;\log_3\left(\frac{16}{36}\right) + 2

    . . . . . . . . . . . . . . . . . . =\;\log_3\left(\frac{4}{9}\right) + 2

    . . . . . . . . . . . . . . . . . . =\; \log_3(4) - \underbrace{\log_3(9)}_{\text{This is 2}} + 2

    . . . . . . . . . . . . . . . . . . =\;\log_3(4) - 2 + 2

    . . . . . . . . . . . . . . . . . . =\;\log_3(4)

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  4. #4
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    Thank you so much Prove It, and Soroban! That was a huge help!

    But, just to clear up one point...Soroban, could you please clarify why log3 (9) becomes 2? Is it because of the "logarithm of the base" rule?

    Cheers,
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  5. #5
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    Quote Originally Posted by BinaryBoy View Post
    ... please clarify why log3 (9) becomes 2?
    let x = \log_3{9}

    then 3^x = 9

    x = 2

    or ...

    \log_3{9} = \log_3(3^2) = 2\log_3{3} = 2 \cdot 1 = 2
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  6. #6
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    Thanks for clarifying, skeeter.
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