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Math Help - eksponential funktion

  1. #1
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    eksponential funktion

    Im stuck with some examples - I would be happy to get some help/guidance

    1)(0,16)x = 5/2
    2) 2x+3*2x-3 =22
    3) 5*0,22-x = (1/25)2x-1
    4)(1/3)2-x -6*9x-4/2 +2*3x-6=29
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  2. #2
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    Re: eksponential funktion

    Hello, skaters!

    These are not easy ones . . . Here is some help.


    1)\;(0.16)^x \:=\:\tfrac{5}{2}

    The left side is: . (0.16)^x \:=\:\left(\tfrac{16}{100}\right)^x \;=\;\left(\tfrac{4}{25}\right)^x \;=\; \left[\left(\tfrac{2}{5}\right)^2\right]^x \;=\;\left(\tfrac{2}{5}\right)^{2x}

    The right side is: . \tfrac{5}{2} \;=\;\left(\tfrac{2}{5}\right)^{-1}

    The equation becomes: . \left(\tfrac{2}{5}\right)^{2x} \;=\;\left(\tfrac{2}{5}\right)^{-1}

    Therefore: . 2x \:=\:-1 \quad\Rightarrow\quad x \:=\:-\tfrac{1}{2}




    3)\;5(0.2)^{2-x} \:=\: \left(\tfrac{1}{25}\right)^{2x-1}

    The left side is: . 5(0.2)^{2-x} \;=\;5\left(\tfrac{1}{5}\right)^{2-x } \;=\;5\left(5^{-1}\right)^{2-x} \;=\;5\cdot 5^{x-2} \;=\;5^{x-1}

    The right side is: . \left(\tfrac{1}{25}\right)^{2x-1} \;=\; \left(5^{-2}\right)^{2x-1} \;=\;5^{-4x+2}

    The equation becomes: . 5^{x-1} \;=\;5^{-4x+2}

    Therefore: . x-1 \:=\:-4x+2 \quad\Rightarrow\quad 5x \:=\:3 \quad\Rightarrow\quad x \:=\:\tfrac{3}{5}




    4)\;\left(\tfrac{1}{3}\right)^{2-x} - 6\left(9^{\frac{x-4}{2}}\right) +2\left(3^{x-6}\right) \:=\:29

    The first term is: . \left(\tfrac{1}{3}\right)^{2-x} \;=\;\left(3^{-1}\right)^{2-x} \;=\;3^{x-2}

    The second term is: . 6\left(9^{\frac{x-4}{2}}\right) \;=\;6\left(3^2\right)^{\frac{x-4}{2}} \;=\;6\left(3^{x-4}\right) \;=\;2\cdot3\,(3^{x-4}) \;=\;2\cdot3^{x-3}

    The equation becomes: . 3^{x-2} - 2\cdot3^{x-3} + 2\cdot3^{x-6} - 29 \:=\:0 . . . ugh!


    I see that x = 6 is a solution.
    Beyond that, you're on your own.
    Thanks from skaters
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: eksponential funktion

    If we continue where Soroban left off with the last one...

    3^{x-2}-\frac{2}{3}3^{x-2}+\frac{2}{81}3^{x-2}=29

    \frac{29}{81}3^{x-2}=29

    3^{x-2}=3^4

    x-2=4

    x=6
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  4. #4
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    Re: eksponential funktion

    Thanks guys!

    I did last one with substitution, it turned very ugly with big nummbers(still I managed ). Your method looks more suitable
    3^{x-2}-\frac{2}{3}3^{x-2}+\frac{2}{81}3^{x-2}=29

    \frac{29}{81}3^{x-2}=29
    But I cant get what u do with -\frac{2}{3}
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: eksponential funktion

    3^{x-2}-\frac{2}{3}3^{x-2}+\frac{2}{81}3^{x-2}=29

    Factor the left side:

    3^{x-2}\left(1-\frac{2}{3}+\frac{2}{81} \right)=29

    Get common denominator within parentheses:

    3^{x-2}\left(\frac{81-54+2}{81} \right)=29

    \frac{29}{81}3^{x-2}=29

    \frac{81}{29}\cdot\frac{29}{81}3^{x-2}=29\cdot\frac{81}{29}

    3^{x-2}=81=3^4
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