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Thread: Solving Log expressions

  1. #1
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    Solving Log expressions

    i am lost on what to do with these two questions:
    they are more intense than other ones we have even looked at in class

    solve for x:

    8^6x-6 = (1/16)^8x-2
    x ='s ___ ??

    and another similar one:

    2^2x+9 =3^x-44
    x='s ___ in terms of logarithms or to four decimal places.

    any guidance would be great!
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  2. #2
    Moo
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    Hello,

    Is it

    $\displaystyle 8^{6x}-6 = (1/16)^{8x}-2$

    Or

    $\displaystyle 8^{6x-6} = (1/16)^{8x-2}$

    ?


    $\displaystyle 2^{2x}+9 =3^x-44$

    Or

    $\displaystyle 2^{2x+9} =3^{x-44}$

    ?
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  3. #3
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    Hello, bharriga!

    You should use parentheses when writing math.
    I'll have to guess what you meant.


    $\displaystyle 8^{6x-6} \:= \:\left(\frac{1}{16}\right)^{8x-2}$

    We have: .$\displaystyle \left(2^3\right)^{6x-6} \:=\:\left(2^{-4}\right)^{8x-2} \quad\Rightarrow\quad 2^{18x-18} \:=\:2^{-3x+8}$

    Hence: .$\displaystyle 18x-18 \:=\:-32x+8\quad\Rightarrow\quad 50x \:=\:26\quad\Rightarrow\quad\boxed{ x \:=\:\frac{13}{25}}$



    $\displaystyle 2^{2x+9} \:=\:3^{x-44}$

    Take logs: .$\displaystyle \ln\left(2^{2x+9}\right) \:=\:\ln\left(3^{x-44}\right) \quad\Rightarrow\quad (2x+9)\!\cdot\!\ln(2) \:=\:(x-44)\!\cdot\!\ln(3)$

    . . $\displaystyle 2x\cdot\ln(2) + 9\cdot\ln(2) \:=\:x\cdot\ln(3) - 44\cdot\ln(3) \quad\Rightarrow\quad x\cdot\ln(3) - 2x\cdot\ln(2) \:=\:9\cdot\ln(2) - 44\cdot\ln(3) $


    Factor: .$\displaystyle x\,[\ln(3) - 2\!\cdot\!\ln(2)] \:=\:9\!\cdot\!\ln(2) - 44\!\cdot\!\ln(3) $


    Therefore . $\displaystyle \boxed{x \;=\;\frac{9\!\cdot\!\ln(2) - 44\!\cdot\!\ln(3)}{\ln(3) - 2\!\cdot\!\ln(2)} \;\approx\;-189.7138}$

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