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Math Help - Log fraction with no calculator

  1. #1
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    Log fraction with no calculator

    How would I solve \log_3 {\frac94} + \log_3 4 without using a calculator.

    This is far as I got:

    \frac{\log 9/4}{\log 3} + \frac{\log 4}{\log 3}

    Thanks!
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  2. #2
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    Quote Originally Posted by desiderius1 View Post
    How would I solve \log_3 {\frac94} + \log_3 4 without using a calculator.

    This is far as I got:

    \frac{\log 9/4}{\log 3} + \frac{\log 4}{\log 3}

    Thanks!
    \log_3 \left({\frac94}\right) + \log_3 (4) = \log_3(9)-\log_3(4)+\log_3(4)

    Simplify! Keep in mind that \log_3(9)=2
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  3. #3
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    Thank you.

    I have another log question. Cannot use calculator.

    \log_6 8 + \log_6 9 - \log_6 2

    So, should I add or subtract first?

    if I add \log_6 (8*9) I get \log_6 (72) which I can't figure out how to simplify any further. If I subtract \log_6 \frac{9}{2}, what would I do next?

    Thanks for helping.
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  4. #4
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    Ah, ok I figured it out.

    log_6 8 + log_6 9 = log_6 (8*9) = log_6 (72)
    log_6 \frac{72}{2} = log_6 36<br />
= log_6 6^2 = 2
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  5. #5
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    more log questions:

    log_6 4 - log_6 3 - log_6 8

    so far I've done:

    log_6 4 - log_6 8 = log_6 {\frac48} = log_6 {\frac12}
    log_6{\frac12} - log_6 3
    Stuck here.


    Other problems...
    log_7 \frac{3}{147}

    log_2(log_4 2 + log_4 8)^3

    3(4log_4 12-2log_4 9)

    thanks!~
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  6. #6
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    Quote Originally Posted by desiderius1 View Post
    more log questions:

    log_6 4 - log_6 3 - log_6 8

    so far I've done:

    log_6 4 - log_6 8 = log_6 {\frac48} = log_6 {\frac12}
    log_6{\frac12} - log_6 3
    Stuck here.
    Re-write the difference of logs as a log of a quotient:

    \log_6\left(\frac12\right) - \log_6( 3)= \log_6\left(\dfrac{\frac12}{3}\right)=\log_6 \left(\frac16 \right) = \log_6(6^{-1})

    Continue!

    Other problems...
    log_7 \frac{3}{147}

    log_2(log_4 2 + log_4 8)^3

    3(4log_4 12-2log_4 9)

    thanks!~
    To help you sufficiently I must know what you have done so far and where you have difficulties doing these questions.
    Please show your work.
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  7. #7
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    Quote Originally Posted by desiderius1 View Post
    Other problems...
    log_7 \frac{3}{147}
    Try log7 ((147/3)^-1).

    Edit:
    Quote Originally Posted by desiderius1 View Post
    <br />
log_2(log_4 2 + log_4 8)^3<br />
    log4 2 <=> 4^x = 2 and log4 8 <=> 4^y = 8. Solve x and y first. Makes it a lot easier.
    Last edited by Dimple; April 14th 2010 at 07:52 AM.
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  8. #8
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    Quote Originally Posted by desiderius1 View Post
    more log questions:

    log_6 4 - log_6 3 - log_6 8

    so far I've done:

    log_6 4 - log_6 8 = log_6 {\frac48} = log_6 {\frac12}
    log_6{\frac12} - log_6 3
    Stuck here.


    Other problems...
    log_7 \frac{3}{147}

    log_2(log_4 2 + log_4 8)^3

    3(4log_4 12-2log_4 9)

    thanks!~
    Note that 3 will cancel: \frac{3}{147} = \frac{1}{49} which is much easier to manipulate
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  9. #9
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    logs

    [quote=desiderius1;493586]more log questions:

    log_6 4 - log_6 3 - log_6 8

    so far I've done:

    log_6 4 - log_6 8 = log_6 {\frac48} = log_6 {\frac12}
    log_6{\frac12} - log_6 3
    Stuck here.


    Other problems...
    log_7 \frac{3}{147}

    log_2(log_4 2 + log_4 8)^3



    Log B2 {logB4 (2) +logB4(8) }^3

    Work inside brackets first
    !/2 +logB4(16/2)=1/2 +2 -1/2 = 2 2^3=8
    Now you have logB2(8)=3


    bjh
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