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Math Help - 3 problems with logarithms

  1. #1
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    3 problems with logarithms

    The following problems are giving me a little grief, mostly because I'm not exactly sure how the solutions should look. Any help is appreciated, I think my biggest problems is not understanding the language of mathematics!

    Problem 1 Solved

    If \log_{10}2=a and \log_{10}3=b, find the exact solution to the equation 12^{x+2}=18^{x-3} in terms of a and b.

    Problem 2 Solved

    Let r=\log_{b}\frac{8}{45} and s=\log_{b}\frac{135}{4}. Find an ordered pair of integers (m, n) such that \log_{b}\frac{32}{5}=mr+ns.

    Problem 3

    Given \log_{18}6=a, find log_{18}16 in terms of a.
    Last edited by mikedwd; October 26th 2008 at 08:19 AM.
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  2. #2
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    This will give you a start.
    \begin{gathered}<br />
  12^{x + 2}  = \left( {2^2 } \right)^{x + 2} \left( 3 \right)^{x + 2}  = \left( {2^{2x + 4} } \right)\left( {3^{x + 2} } \right) \hfill \\<br />
  \log \left( {12^{x + 2} } \right) = \left( {2x + 4} \right)\log (2) + \left( {x + 2} \right)\log (3) \hfill \\ <br />
\end{gathered}
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  3. #3
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    Wow that's a huge help!

    thanks a lot I'm now doing the same thing to the other side of the equation I'll post back with my answer!
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  4. #4
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    I got this far on my answer:

    a(2x+4)+b(x+2)=a(x-3)+b(2x-6)

    But I'm not sure how I could isolate a and b...

    I got there just by replace log2 and log3 with a and b on what Plato did, then did the same method that he did on the second expression, 18^{x-3}, and again replaced log2 and log3 with a and b and came up with the equality above.
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  5. #5
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    So does anyone know how I can finish the first problem (or begin the second 2)?

    Did I set up the first one wrong, or do I just not know how to isolate them?
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  6. #6
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    Continue the first.
    \begin{gathered}<br />
  2ax + 4a + bx + 2b = ax - 3a + 2bx - 6b \hfill \\<br />
  ax - bx =  - 7a - 8b \hfill \\<br />
  x = \frac{{ - 7a - 8b}}<br />
{{a - b}} \hfill \\ <br />
\end{gathered}
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  7. #7
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    ugh....

    I was looking at it all wrong I don't even know why I didn't do that, anyway thanks
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  8. #8
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    I wonder if I'm beginning Problem 2 completely wrong

    I'm not exactly sure how to set it up
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  9. #9
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    \begin{gathered}<br />
  r = 3\log (2) - 2\log (3) - \log (5) \hfill \\<br />
  s = \log (5) + 3\log (3) - 2\log (2) \hfill \\<br />
\hfill \\<br />
  3r = 9\log (2) - 6\log (3) - 3\log (5) \hfill \\<br />
  2s = 2\log (5) + 6\log (3) - 4\log (2) \hfill \\<br />
  3r + 2s = 5\log (2) - \log (5) = \log \left( {\frac{{32}}<br />
{5}} \right) \hfill \\ <br />
\end{gathered}
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  10. #10
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    Does anyone have any kind of hint for me of what I have to do to solve the third problem?
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  11. #11
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    Quote Originally Posted by mikedwd View Post

    Problem 3

    Given \log_{18}6=a, find log_{18}16 in terms of a.
    Having \log _{18}6=a, it's not hard to prove that \log _{2}3=\frac{1-a}{2a-1}, hence

    \log _{18}16=\frac{4}{\log _{2}18}=\frac{4}{1+2\cdot \dfrac{1-a}{2a-1}}=8a-4,

    as required.
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  12. #12
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    Thanks a lot! It seems so clear now that I see it, but this should help me a lot with more problems to come.

    I have one more question, though.

    On this problem:

    Find all real numbers x for which \frac{3^{\sqrt{12x}}+3}{4}=3^{\sqrt{3x}}.

    I did as follows:

    3^{\sqrt{12x}}=4*3^{\sqrt{3x}}-3

    \frac{3^{\sqrt{12x}}}{3^{\sqrt{3x}}}=1

    3^{\sqrt{12x}-\sqrt{3x}}=1

    3^{\sqrt{9x}}=1

    x=0

    Is this the only solution or are there others? This was one of two questions were it asked for "all real numbers", and the other one had multiple solutions.

    I think that 0 is the only solution, but how can I be sure?
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