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Math Help - Testing solutions in logarithmic equations

  1. #1
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    Unhappy Testing solutions in logarithmic equations

    Hello I'm a newbie. I need help with my homework. I have ADD bad. I don't understand this section Could someone please show how to work these problems?

    In Ex. 1-8, determine whether each x-value is a solution of the equation.


    3. 3e^x+2=75
    (b) -2 + ln25
    (c) x approx 1.2189

    7. ln(x-1)=3.8
    (b) x approx 45.7012
    (c) x=1 + ln 3.8
    Last edited by mr fantastic; April 22nd 2009 at 10:04 PM. Reason: Distributing questions from another thread, corrected a typo
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  2. #2
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    Quote Originally Posted by asimon2008 View Post
    Hello I'm a newbie. I need help with my homework. I have ADD bad. I don't understand this section Could someone please show how to work these problems?

    In Ex. 1-8, determine whether each x-value is a solution of the equation.


    3. 3e^x+2=75
    (b) -2 + ln25
    (c) x approx 1.2189

    7. ln(x-1)=3.8
    (b) x approx 45.7012
    (c) x=1 + ln 3.8
    A) There are two different ways to do this question:
    1. Plug in the given results and check if the equation becomes a true statement.
    2. Solve the equation for x and check if you've got one of the given results.

    B) to #3 (way #2.):
    I assume that the original equation reads: 3e^(x+2)=75 . If so:

    3e^{x+2}=75~\implies~e^{x+2}=25~\implies~x+2=\ln(2  5)~\implies~ x=-2+\ln(25) \approx 1.218875825...

    Therefore both answers are correct.


    to #7 (way #1):Plug in x = 1 + ln(3.8):

    \ln(1 + \ln( 3.8) - 1) = 3.8~\implies~\ln(\ln(3.8)) \neq 3.8

    Plug in x = 45.7012:

    \ln(45.7012 - 1) = \ln(44.7012)\approx 3.8000000347...

    Therefore x = 45.7012 is an approximate solution of the equation.
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