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Math Help - logarithms help

  1. #1
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    logarithms help

    Solve the simultaneous equations  4^x=5^y and 2(4^x)=7^y.

    I have tried substituting 5^y into 2(4^x)=7^y and also changing  4^x=5^y into y=log_{5}4^x. I seem to be going on and on without really going anywhere with my working.
    Could really do with some help. Thanks
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  2. #2
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    Quote Originally Posted by greghunter View Post
    Solve the simultaneous equations  4^x=5^y and 2(4^x)=7^y.

    I have tried substituting 5^y into 2(4^x)=7^y and also changing  4^x=5^y into y=log_{5}4^x. I seem to be going on and on without really going anywhere with my working.
    Could really do with some help. Thanks
    You have 2(5^y)=7^y, so taking logs (to any base) you have:

    \log(2)+y\log(5)=y\log(7)

    hence:

    y= ...

    CB
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  3. #3
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    Quote Originally Posted by greghunter View Post
    Solve the simultaneous equations  4^x=5^y and 2(4^x)=7^y.

    I have tried substituting 5^y into 2(4^x)=7^y and also changing  4^x=5^y into y=log_{5}4^x. I seem to be going on and on without really going anywhere with my working.
    Could really do with some help. Thanks
    2\left(4^x\right) = 7^y and 4^x = 5^y.

    Therefore

    2\left(5^y\right) = 7^y

    2 = \frac{7^y}{5^y}

    2 = \left(\frac{7}{5}\right)^y

    \ln{2} = \ln{\left(\frac{7}{5}\right)^y}

    \ln{2} = y\ln{\frac{7}{5}}

    \ln{2} = y(\ln{7} - \ln{5})

    y = \frac{\ln{2}}{\ln{7} - \ln{5}}.


    Can you solve for x?
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  4. #4
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    Thanks, I didn't realise you could stick a log infront but it does make sense
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  5. #5
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    Quote Originally Posted by greghunter View Post
    Solve the simultaneous equations  4^x=5^y and 2(4^x)=7^y.

    I have tried substituting 5^y into 2(4^x)=7^y and also changing  4^x=5^y into y=log_{5}4^x. I seem to be going on and on without really going anywhere with my working.
    Could really do with some help. Thanks

    Divide the second equation by the first giving 2 = (7^y)/(5^y)
    then 2 = (7/5)^y or 2 = (1.4)^y

    Solve for y = ln 2 / ln 1.4

    Substitute into the first equation for y
    4^(x) = 5 ^(ln 2/ ln 1.4)

    x ln 4 = (ln2/ln1.4) (ln 5); change ln 4 to ln 2^2 = 2 ln 2

    x = (ln 1.4) (ln 5)/(ln 2)
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