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Math Help - Logarithmic Equations

  1. #1
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    Logarithmic Equations

    Test time again, struggling on a few questions on the practice test and simply looking for some assistance on tackling the equations!
    Question 1:
    Consider the following logarithmic equations. Solve for x.

    A)


    B)

    Question 2:
    Consider the following exponential equations. Solve for x.



    Thanks for all your help and support!
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  2. #2
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    Re: Logarithmic Equations

    A)

    2 log^2(x)-5 log(x)-3 = 0
    Substitute u = log(x):
    2 u^2-5 u-3 = 0
    Solve the quadratic equation by completing the square:

    Divide both sides by 2:
    u^2-(5 u)/2-3/2 = 0
    Add 3/2 to both sides:
    u^2-(5 u)/2 = 3/2
    Add 25/16 to both sides:
    u^2-(5 u)/2+25/16 = 49/16
    Factor the left hand side:
    (u-5/4)^2 = 49/16
    Take the square root of both sides:
    abs(u-5/4) = 7/4
    Eliminate the absolute value:
    u-5/4 = -7/4 or u-5/4 = 7/4
    Add 5/4 to both sides:
    u = -1/2 or u-5/4 = 7/4
    Substitute back for u = log(x):
    log(x) = -1/2 or u-5/4 = 7/4
    Cancel logarithms by taking exp of both sides:
    x = 1/sqrt(e) or u-5/4 = 7/4
    Add 5/4 to both sides:
    x = 1/sqrt(e) or u = 3
    Substitute back for u = log(x):
    x = 1/sqrt(e) or log(x) = 3
    Cancel logarithms by taking exp of both sides:
    x = 1/sqrt(e) or x = e^3
    Now test that these solutions are appropriate by substitution into the original equation:
    Check the solution x = 1/sqrt(e):
    2 log^2(x)-5 log(x)-3 => -3-5 log(1/sqrt(e))+2 log^2(1/sqrt(e)) = 0
    So the solution is correct.
    Check the solution x = e^3:
    2 log^2(x)-5 log(x)-3 => -3-5 log(e^3)+2 log^2(e^3) = 0
    So the solution is correct.
    Thus, the solutions are:
    x = 1/sqrt(e) or x = e^3

    B)

    (log(x))/(log(2)) = (log(18))/(log(2))-(log(6))/(log(2))
    Divide both sides by 1/(log(2)):
    log(x) = log(2) ((log(18))/(log(2))-(log(6))/(log(2)))
    Cancel logarithms by taking exp of both sides:
    x = 3
    Now test that this solution is appropriate by substitution into the original equation:
    Check the solution x = 3:
    (log(x))/(log(2)) => (log(3))/(log(2)) = (log(3))/(log(2)) ~~ 1.58496
    (log(18))/(log(2))-(log(6))/(log(2)) => -(log(6))/(log(2))+(log(18))/(log(2)) = (log(3))/(log(2)) ~~ 1.58496
    So the solution is correct.
    Thus, the solution is:
    x = 3

    Basically, if you see a quadratic pattern like with A), e.g. in the form ax^2+x+c, where c=some constant, then solve using quadratic equation or completing square. Best to make sure you have your log rules down too:
    Mathwords: Logarithm Rules

    Hope that helps? Best of luck! ("problem 2's" picture appears to not be working...not displaying correctly?)
    Last edited by MrCryptoPrime; April 18th 2012 at 02:51 PM.
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  3. #3
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    Re: Logarithmic Equations

    Quote Originally Posted by Prentz View Post
    B)
    \log_2{x} = \log_2{18} - \log_2{6}

    \log_2{x} = \log_2\left(\frac{18}{6}\right)

    \log_2{x} = \log_2\left{3}

    x = 3

    ... if you meant to post more after this equation (question 2 ?), it did not show.
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  4. #4
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    Re: Logarithmic Equations

    Thanks for the help guys, is there an easier way to express question 1? It confused me slightly.
    And as for question 2:
    Consider the following exponential equations. Solve for x.
    Logarithmic Equations-_large_-_frac-9-2x-1-3-3x-2-27-x-2-4-81-3-.gif
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  5. #5
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    Re: Logarithmic Equations

    Any help you can give on that question 2 and clearing up question 1 would be great!
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  6. #6
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    Re: Logarithmic Equations

    Do you know anything about the "laws of exponents"?

    a^xa^y= a^{x+y}
    (a^x)^y= a^{xy}
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  7. #7
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    Re: Logarithmic Equations

    Laws of exponents.
    Last edited by highvoltage; April 20th 2012 at 06:11 AM.
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  8. #8
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    Re: Logarithmic Equations

    Logarithmic Equations-expon.jpg
    Last edited by highvoltage; April 19th 2012 at 07:36 PM.
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  9. #9
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    Re: Logarithmic Equations

    Quote Originally Posted by Prentz View Post

    A)
    Yet another solution...why?

    2(\log x)^2-5\log x -3=0

    (2\log x +1)(\log x-3)=0

    \log x = -\frac{1}{2} or \log x =3

    x=e^{-1/2} or x=e^3
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