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Math Help - Some Basic Logarithims Questions!

  1. #1
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    Some Basic Logarithims Questions!

    Ok guys, just brushing up on some review questions and a few have stumped me. Appreciate it if you can help me out WITH HOW you solved the question. Thanks! Now the questions...

    1) \frac {\log_{11} 64}{\log_{11} 16}

    **This one I thought was really easy but I got it wrong... So, If the base is the same (11), can't I just divide 64/16=4? Or, wait, can this be written as:

    {\log_{11} 64}- {\log_{11} 16}

    So {\log_{11} 48} would be my answer?

    2) Simplify
    <br />
ln(a^2-b^2)-ln(a+b)<br />

    as much as possible.

    ** This again is a difference question so I can rewrite this as a quotient and solve? Right track?

    3) If <br />
x=e^y. write ye^y as a simple function of x.

    ** I suck at word problems or interpreting you can say. Kind of lost on this one but I am thinking, re write this as an exponential function? Not sure, just throwing it out there!
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  2. #2
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    Quote Originally Posted by mvho View Post
    Ok guys, just brushing up on some review questions and a few have stumped me. Appreciate it if you can help me out WITH HOW you solved the question. Thanks! Now the questions...

    1) \frac {\log_{11} 64}{\log_{11} 16}

    **This one I thought was really easy but I got it wrong... So, If the base is the same (11), can't I just divide 64/16=4? Or, wait, can this be written as:

    {\log_{11} 64}- {\log_{11} 16}

    So {\log_{11} 48} would be my answer?
    No, neither method is right.
    \frac{\log a}{\log b} \neq \log a - \log b. You're confusing with
    \log \left(\frac{a}{b}\right) = \log a - \log b.

    Instead, do this:
    \begin{aligned}<br />
\frac{\log_{11} 64}{\log_{11} 16} &= \frac{\log_{11} 4^3}{\log_{11} 4^2} \\<br />
&= \frac{3\log_{11} 4}{2\log_{11} 4} \\<br />
&= \frac{3}{2}<br />
\end{aligned}


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  3. #3
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    Quote Originally Posted by mvho View Post
    2) Simplify
    <br />
ln(a^2-b^2)-ln(a+b)<br />

    as much as possible.

    ** This again is a difference question so I can rewrite this as a quotient and solve? Right track?
    Yes:

    \begin{aligned}<br />
\ln (a^2-b^2)- \ln(a+b) &= \ln \left(\frac{a^2-b^2}{a + b}\right) \\<br />
&= \ln \left(\frac{(a + b)(a - b)}{a + b}\right) \\<br />
&= \ln (a - b)<br />
 \end{aligned}
    This is all assuming, of course, that everything inside the logarithms are positive numbers.


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  4. #4
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    Quote Originally Posted by mvho View Post

    3) If <br />
x=e^y. write ye^y as a simple function of x.

    x=e^y \Rightarrow y=ln(x)

    now ye^y = ln(x)e^{ln(x)} = x\times ln(x)
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  5. #5
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    Wow, Thanks alot guys. I feel silly now seeing how simple it is. For question #1, find the same base and solve.

    For question #2, I should've caught the difference of squares. It seems obvious now that you have solved it. LOL
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