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Math Help - Logarithms

  1. #1
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    Logarithms

    Please help me solve the following problems:

    1.) log_4(3x-2)^{1/5} = 1/10

    2.) ln(x) + ln(x+6) = 0.5(ln 9)

    3.) 30,000 = \frac {50,500}{1 + 100e^{-0.7x}}

    Your help is greatly appreciated.
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by currypuff View Post
    Please help me solve the following problems:

    1.) log_4(3x-2)^{1/5} = 1/10

    2.) ln(x) + ln(x+6) = 0.5(ln 9)

    3.) 30,000 = \frac {50,500}{1 + 100e^{-0.7x}}

    Your help is greatly appreciated.
    1.) log_4(3x-2)^{1/5} = 1/10

    (3x-2)^{1/5}=\left(4\right)^{\frac{1}{10}}

    (3x-2)=\left(4\right)^{\frac{5}{10}}

    3x-2=\sqrt{4}

    (3x-2)^2=4

    9x^2-12x+4=4

    9x^2=12x

    x \ne 0

    9x=12

    x=\frac{12}{9}
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  3. #3
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    Hello, currypuff!

    1)\;\;\log_4(3x-2)^{\frac{1}{5}} \;=\; \frac{1}{10}

    We have: . \frac{1}{5}\log_4(3x-2) \:=\:\frac{1}{10}

    Multiply by 5: . \log_4(3x-2)\:=\:\frac{1}{2}

    Exponentiate: . 3x-2\;= \;4^{\frac{1}{2}}\quad \Rightarrow\quad 3x-2 \;=\;2

    Therefore:. 3x\;=\;4\quad\Rightarrow\quad\boxed{x \:=\:\frac{4}{3}}




    2)\;\;\ln(x) + \ln(x+6) \:= \:\frac{1}{2}\ln 9

    We have: . \ln[x(x+6)] \:=\:\ln(9^{\frac{1}{2}})\quad\Rrightarrow\quad \ln(x^2+6x) \:=\:\ln(3)<br />

    "Un-log" both sides: . x^2-6x\:=\:3\quad\Rightarrow\quad x^2 -6x - 3\:=\:0

    Quadratic Formula: . x \;=\;\frac{-6 \pm\sqrt{48}}{2} \;=\;-3 \pm2\sqrt{3}

    The negative root is extraneous.

    Therefore: . \boxed{x \;=\;2\sqrt{3} - 3}




    3)\;\;\frac{50,500}{1 + 100e^{-0.7x}} \;=\;30,000

    Divide by 50,500: . \frac{1}{1 + 100e^{-0.7x}} \:=\:\frac{60}{101}

    "Cross-multiply": . 101 \:=\:60 + 6000e^{-0.7x} \quad\Rightarrow\quad 41 \;=\;6000^{-0.7x}

    Multiply by e^{0.7x}\!:\;\;41e^{0.7x}\;=\;6000 \quad\Rightarrow\quad e^{0.7x} \;=\;\frac{6000}{41}

    Therefore: . 0.7x \;=\;\ln\left(\frac{6000}{41}\right) \quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{0.7}\ln\left(\frac{6000}{41}\right)}<br /> <br />
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