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    Hi i'm trying to solve for x by using log but I'm not sure how to do it

    3^x = \frac{-1+\sqrt 73}{4}

    I rearranged it to xln3 = \frac{ln-1\sqrt 73}{4}

    that doesn't look correct, can someone please tell me what I'm doing wrong?

    thanks
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    (Original post by HelloGoodbye)
    Hi i'm trying to solve for x by using log but I'm not sure how to do it

    3^x = \frac{-1+\sqrt 73}{4}

    I rearranged it to xln3 = ln\frac{-1+\sqrt 73}{4}

    that doesn't look correct, can someone please tell me what I'm doing wrong?

    thanks
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    That is fine
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    (Original post by TenOfThem)
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    That is fine
    thanks! and I'm trying to do this simultaneous equation

    2^x - 3^y = -1
    2^(x-1) - 3^(y-1) = -1

    would I change the first equation to log like this:

    xln2 - yln3 = -1
    I'm not sure what to do after that?
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    (Original post by HelloGoodbye)
    thanks! and I'm trying to do this simultaneous equation

    2^x - 3^y = -1
    2^(x-1) - 3^(y-1) = -1

    would I change the first equation to log like this:

    xln2 - yln3 = -1
    I'm not sure what to do after that?
    I don't know what you've tried to do there, and it certainly isn't valid!!

    I would use the first equation to rewrite 3^y in terms of 2^x (or vice versa) and then substitute into the second equation.
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    (Original post by HelloGoodbye)
    thanks! and I'm trying to do this simultaneous equation

    2^x - 3^y = -1
    2^(x-1) - 3^(y-1) = -1

    would I change the first equation to log like this:

    xln2 - yln3 = -1
    I'm not sure what to do after that?
    No

    You cannot log individual parts, plus you have not changed the -1

    Follow davros' advice
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    (Original post by davros)
    I don't know what you've tried to do there, and it certainly isn't valid!!

    I would use the first equation to rewrite 3^y in terms of 2^x (or vice versa) and then substitute into the second equation.
    ok thanks so I've rearraged the equation to give me 2^x = 3^y-1 and sub'd that into equation 2:

    (3^y-1)^-2 - 3^(y-1) + 1=0

    \frac{1}{(3^y-1)^2} - \frac{1}{3^y} +1=0

    Am I doing that right?
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    (Original post by HelloGoodbye)
    ok thanks so I've rearraged the equation to give me 2^x = 3^y-1 and sub'd that into equation 2:

    (3^y-1)^-2 - 3^(y-1) + 1=0

    \frac{1}{(3^y-1)^2} - \frac{1}{3^y} +1=0

    Am I doing that right?
    No

    Where has the 1/3^y come from and where has the -2 power come from
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    (Original post by HelloGoodbye)
    ok thanks so I've rearraged the equation to give me 2^x = 3^y-1 and sub'd that into equation 2:

    (3^y-1)^-2 - 3^(y-1) + 1=0

    \frac{1}{(3^y-1)^2} - \frac{1}{3^y} +1=0

    Am I doing that right?
    No, not at all

    2^{x-1} = \dfrac{2^x}{2}

    and

    3^{y-1} = \dfrac{3^y}{3}

    You seem to have complicated things unnecessarily.
 
 
 
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