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Math Help - Solving Equations..please Help!

  1. #1
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    Solving Equations..please Help!

    Hi, I have the following equations that I am really confused on..just wondering if someone can help me out with them..thanks!

    1) 2e^7x+2=6


    2) 1n (3x-4) + 1n (x-3)-1n 4=0
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  2. #2
    Senior Member topher0805's Avatar
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    2e^{7x}+2=6

    Subtract 2 from both sides:

    2e^{7x}=4

    Divide both sides by 2:

    e^{7x}=2

    Take the natural log of both sides:

    \ln e^{7x}=\ln 2

    Recall this law of logarithms:

    log (x)^a = a\cdot log (x)

    Apply it:

    7x\cdot \ln e = \ln 2

    The natural log of e is 1, so we have:

    7x = \ln 2

    Divide both sides by 7;

    x = \frac {\ln 2}{7}

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  3. #3
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    For problem 1 rearrange your equation so that your exponentiated factor ( e^{7x}) is by itself (no coefficient) then treat each side of the equation like a variable and plug each into the inverse of e^y . (hint: You should be using the fact that \ln(e^y) = y.)

    For problem 2 observe that \ln(a)+\ln(b) = \ln(ab) and as an immediate consequence \ln(c) - \ln(d) = \ln(c/d). Also note that e^{\ln(y)} = y
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  4. #4
    Senior Member topher0805's Avatar
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    \ln (3x-4) + \ln (x-3)-\ln 4=0

    Recall this law of logarithms:

    \log (a) + \log (b) = \log (ab)

    Apply it:

    \ln \left((3x-4)\cdot (x-3)\right) - \ln 4 = 0

    <br />
\ln (3x^2 - 13x + 12) - \ln 4 = 0

    <br />
\ln (3x^2 - 13x + 12) = \ln 4

    e^{\ln 4} = 3x^2 - 13x + 12

    Recall that:

    e^{\ln x} = x

    Apply it:

    3x^2 - 13x + 12 = 4

    3x^2 - 13x + 8 = 0

    Solve for x.
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