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Math Help - Logarithmic equation

  1. #1
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    Logarithmic equation

    Solve for x:
    \log_{2008}\left(x-3\right)+\log_{2009}\left(x-3\right)=3-\lg\left(x^5-24\right)
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  2. #2
    Super Member Aryth's Avatar
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    Well, we are given:

    log_{2008}{(x - 3)} + log_{2009}{(x - 3)} = 3 - log(x^5 - 24)

    Well, let's deal with the right side first, they are currently in different bases, therefore we must change the bases by using the change of base formula:

    log_a{b} = \frac{log{b}}{log{a}}

    Where we are using base 10 because the right side has a base 10 logarithm.

    Therefore we get:

    \frac{1}{log{2008}}log{(x - 3)} + \frac{1}{log{2009}}log{(x - 3)} = 3 - log{(x^5 - 24)}

    Where:

    log{2008} = 3.203

    and

    log{2009} = 3.203

    (They are very close together, where the first difference is in the ten thousandths place.)

    So we get:

    0.31223log{(x - 3)} + 0.31223log{(x - 3)} = 0.31223(log{(x - 3)^2})

    0.62446log{(x - 3)} = 3 - log{(x^5 - 24)}

    There's really not an easy way to solve this, so I'm going to make the following approximation:

    0.62446 \approx \frac{2}{3}

    log{(x - 3)^{\frac{2}{3}}} = 3 - log(x^5 - 24)

    log{(x - 3)^{\frac{2}{3}}} + log(x^5 - 24) = 3

    log{(x - 3)^{\frac{2}{3}}(x^5 - 24)} = 3

    (x - 3)^{\frac{2}{3}}(x^5 - 24) = 10^3

    \sqrt[3]{x^5 - 24} = \frac{10}{(x - 3)^2}

    (x^5 - 24) = \frac{1000}{(x - 3)^5}

    (x^5 - 24)(x + (-3))^5 = 1000

    (x^5 - 24)(x^5 - 15x^4 + 90x^3 - 270x^2 + 405x - 243) = 1000

    x^{10} - 15x^9 + 90x^8 - 270x^7 + 405x^6 - 267x^5 + 360x^4 - 2160^3 + 6480x^2 - 9720x + 4832 = 0

    You get:

    x = 4

    To verify:

    <br />
\frac{\log{1}}{log{2008}}+ \frac{\log{1}}{log{2009}} = 3 - \log\left(1000\right)<br />

    0 + 0 = 3 - 3 = 0

    And there you go.
    Last edited by Aryth; March 17th 2008 at 06:05 PM.
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  3. #3
    Senior Member topher0805's Avatar
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    <br />
\sqrt[3]{x^5 - 24} = \frac{10}{(x - 3)^2}<br />
    How did you get from that to the step below? It just doesn't make any sense...

    <br />
(x^5 - 24) = \frac{1000}{(x - 3)^5}<br />
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  4. #4
    Super Member Aryth's Avatar
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    All I did was cube both sides, and since cubing anything keeps the sign it currently has, I didn't need a plus or minus, but you are right, I was a power off:

    <br />
\sqrt[3]{x^5 - 24} = \frac{10}{(x - 3)^2}<br />

    (\sqrt[3]{x^5 - 24})^3 = \left(\frac{10}{(x - 3)^2}\right)^3

    Evaluated, it equals:

    (x^5 - 24) = \frac{10^3}{((x - 3)^2)^3} = \frac{1000}{(x - 3)^6}

    But nonetheless, 4 is still the answer.
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  5. #5
    Senior Member topher0805's Avatar
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    It's interesting that you get the correct answer despite making an error...
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  6. #6
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    Just to clarify: x=4 is a solution. There can't be any other solutions because the LHS is strictly increasing but RHS is strictly decreasing because the fuction \log_bx is stricly increasing if b>1 ( x>0).
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  7. #7
    Super Member Aryth's Avatar
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    With equations have orders that high, there are very few differences. As a matter of fact, with the equation that I had before(the wrong one), it had all of its roots at (4,0).

    Just for fun:

    (x + (-3))^6

    = x^6 - 18x^5 + 135x^4 - 540x^3 + 1214x^2 - 1458x + 729

    Now if we take (x^5 - 24) and multiply we get:

    x^{11} - 18x^{10} + 135x^9 - 540x^8 + 1214x^7 - 1482x^6 + 1161x^5 - 3240x^4 + 12960x^3  \ - 29136x^2 + 34992x - 17496 = 1000

    x^{11}- 18x^{10} + 135x^9 - 540x^8 + 1214x^7 - 1482x^6 + 1161x^5 - 3240x^4 + 12960x^3  \ -29136x^2 + 34992x - 18496 = 0

    And that also yields 4 as it's only zero.
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