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Math Help - Solving Logarithmic Equations

  1. #1
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    Unhappy Solving 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 73-92, solve the logarithmic equation algebraically. Round the result to three decimal places. Verify your answer using a graphing utility.

    81. 7 log4(0.6x)=12

    83. ln sq. root(x+2)=1

    89. ln(x+5)=ln(x-1)-ln(x+1)
    Last edited by mr fantastic; April 22nd 2009 at 11:09 PM. Reason: Distributing questions from another thread
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  2. #2
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    Hello, asimon2008!

    Welcome aboard!


    Solve the logarithmic equation algebraically.
    Round the result to three decimal places.

    (81)\;\;7\log_4(0.6x)\:=\:12

    We have: . \log_4(0.6x) \:=\:\tfrac{12}{7}

    . . . . . . . 0.6x \;=\;4^{\frac{12}{7}}

    . . . . . . . x \;=\; \frac{4^{|frac{12}{7}}}{0.6} \;=\;17.9453359 \;\approx\;17.945




    (83)\;\;\ln\sqrt{x+2} \:=\:1

    We have: . \ln\sqrt{x+2} \:=\:1 \quad\Rightarrow\quad \sqrt{x+2} \;=\;e^1

    . . . . . . . x+ 2 \;=\;e^2 \quad\Rightarrow\quad x \;=\;e^2-2\;\approx\;5.389




    (89)\;\;\ln(x+5)\:=\:\ln(x-1)-\ln(x+1)

    We have: . \ln(x+5) \;=\;\ln\left(\frac{x-1}{x+1}\right) \quad\Rightarrow\quad x+5 \;=\;\frac{x-1}{x+1}

    . . . . . . . . (x+5)(x+1) \:=\:x-1 \quad\Rightarrow\quad x^2 + 6x + 5 \;=\;x - 1

    . . . . . . . . x^2 + 6x + 5 \:=\:0\quad\Rightarrow\quad (x+2)(x+3) \:=\:0

    Hence: . x \;=\;\text{-}2,\:\text{-}3

    But both roots are extraneous . . . The equation has no solution.

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