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Math Help - Logarithmic Equation

  1. #1
    Junior Member
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    Smile Logarithmic Equation

    Hi, Could anyone help me solve this...?


    log3 + log3 = 0
    4x+1 3x


    Thanks
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  2. #2
    MHF Contributor
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    Quote Originally Posted by alternative View Post
    Hi, Could anyone help me solve this...?


    log3 + log3 = 0
    4x+1 3x


    Thanks
    hi

    i assume its \log_3(4x+1)+\log_3(3x)=0

    If you are not sure how to generate these codes , click on it to see the codes .

    Using the properties of log , where log a + log b = log (ab)

    \log_3(4x+1)(3x)=0

    Then convert this to indices form ,

    (4x+1)(3x)=3^0=1

    Solve for x then .
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  3. #3
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    Hello, alternative!


    Is this what you meant?


    \log_{4x+1}(3) + \log_{3x}(3) \;=\;0

    Use the Base-Change Formula: . \frac{\ln(3)}{\ln(4x+1)} + \frac{\ln(3)}{\ln(3x)} \;=\;0

    Multiply by \frac{\ln(4x+1)\ln(3x)}{\ln(3)}\!:\quad \ln(3x) + \ln(4x+1) \;=\;0  \quad\Rightarrow\quad \ln(4x+1) \:=\:-\ln(3x)

    . . . . . . . . . . . . . . . . . . . . . \ln(4x+1) \;=\;\ln(3x)^{-1} \quad\Rightarrow\quad \ln(4x+1) \;=\;\ln\left(\frac{1}{3x}\right)


    Expnentiate both sides: . 4x + 1 \:=\:\frac{1}{3x} \quad\Rightarrow\quad 12x^2 + 3x - 1 \:=\:0

    Quadratic Formula: . x \:=\:\frac{-3 \pm\sqrt{57}}{24}


    The negative root, \frac{-3 - \sqrt{57}}{24}, is extraneous.


    The base of a logarithm must be positive.


    Therefore: . x \;=\;\frac{-3 + \sqrt{57}}{24}

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  4. #4
    Junior Member
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    Smile

    Quote Originally Posted by Soroban View Post
    Hello, alternative!


    Is this what you meant?


    Use the Base-Change Formula: . \frac{\ln(3)}{\ln(4x+1)} + \frac{\ln(3)}{\ln(3x)} \;=\;0

    Multiply by \frac{\ln(4x+1)\ln(3x)}{\ln(3)}\!:\quad \ln(3x) + \ln(4x+1) \;=\;0 \quad\Rightarrow\quad \ln(4x+1) \:=\:-\ln(3x)

    . . . . . . . . . . . . . . . . . . . . . \ln(4x+1) \;=\;\ln(3x)^{-1} \quad\Rightarrow\quad \ln(4x+1) \;=\;\ln\left(\frac{1}{3x}\right)


    Expnentiate both sides: . 4x + 1 \:=\:\frac{1}{3x} \quad\Rightarrow\quad 12x^2 + 3x - 1 \:=\:0

    Quadratic Formula: . x \:=\:\frac{-3 \pm\sqrt{57}}{24}


    The negative root, \frac{-3 - \sqrt{57}}{24}, is extraneous.

    The base of a logarithm must be positive.


    Therefore: . x \;=\;\frac{-3 + \sqrt{57}}{24}


    I find no words to thank you......
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