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    can anyone please help me with the following problems?
    any assistant would be greatly appreciated.
    thank you in advance for your knowledge

    Solve:
    log_x(x+6)=2



    log_3x-3(log_3y+2log_3z)


    log x + log(x-15)=2




    2^2x-3*2^x-40=0



    log_4x+log_8x=5


    also, anyone good with story problems? here's one.
    you invest $2000 in an acct that pays 4.3% interest compounded continuously, find to the nearest tenth of a yr when the acct will be worth $10K.
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    The first one goes like this:

    log_x(x + 6) = 2
    => x^2 = x + 6
    x^2 - x - 6 = 0
    (x+2)(x-3) = 0
    x = -2 , 3
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    For the last one:

    log_8 x=1/3 log_2 x
    log_4 x=1/2 log_2 x
    => log_8 x +log_4 x=1/3 log_2 x +1/2 log_2 x=5/6 log_2 x
    => 5/6 log_2 x= 5
    => log_2 x = 6
    => x=2^6=64, which fits
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    Not sure about number 2 but:
    log_3 x -3log_3 y -6log_3 z=log_3 x -log_3 y^3 -log_3 z^6
    =log_3(x/(y^3*z^6))

    3:
    log x + log(x-15)=log x(x-15)
    log x(x-15)=2
    => 10^2=x(x-15)
    x^2-15x=100
    =>x^2-15x-100=0
    =>(x-20)(x+5)=0
    =>x=-5 or 20
    However, 10^x=-5 has no real solutions since x<0. Therefore, x=20.

    4:
    2^2x-3*2^x-40=(2^x-1.5)^2-42.25 (completing the square)
    =>(2^x-1.5)^2=42.25
    =>2^x-1.5=+6.5
    =>2^x=8 or -5
    As 2^x=-5 has no solutions, 2^x=8
    Therefore, x=3
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    And the wordy one:

    If v is value and y is years:

    v=2000*1.043^y
    When v=10000:
    10000=2000*1.043^y
    5=1.043^y
    ln 5=ln 1.043^y
    ln 5=y ln 1.043
    y=ln 5/ln 1.043
    y= 38.2278...years
    =38.2 years (to 1dp)
 
 
 
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