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Math Help - Indices - please help!

  1. #1
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    Indices - please help!

    Hi

    I have come to this stage from solving a logarithm question. Can't seem to proceed. Please help.

    6^z+3^z = 2^z

    3^z.2^z +3^z = 2^z

    3^z = 2^z / (2^z + 2^0)

    How can I find the value of z from here?

    Thank you.

    rgds
    gentsl
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  2. #2
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    3^{z}(2^{z}+1)=2^{z}

    2^{z}+1=(\frac{2}{3})^{z}

    Just looking at this we can see that \frac{1}{2}+1=\frac{3}{2}

    Which would make z = -1.

    You could also make the observation that 1/6+1/3=1/2.
    Last edited by galactus; December 30th 2007 at 03:39 PM.
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  3. #3
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    Thanks Galactus.

    That was done through observation (guess and check), is there any other way to do it?

    gentsl
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  4. #4
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    More observation than guess and check.

    You could play around and see what you can whip up.

    6^{z}+3^{z}=2^{z}

    Divide by 2^{z}:

    3^{z}+(\frac{3}{2})^{z}=1

    3^{z}+3^{z}e^{-zln(2)}=1

    3^{z}(1+e^{-zln(2)})=1

    1+e^{-zln(2)}=\frac{1}{3^{z}}

    1+2^{-z}=3^{-z}

    Now, since 1+2=3, z must equal -1.

    Just playing around, you can always try other things.

    Look into the Lambert W function for tricky exponent problems.
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