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Math Help - Logarithm problem

  1. #1
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    Logarithm problem

    solve x, I have really struggled with this.

    30,000=0.1(60,000)^x+0.9(10,000)^x
    For the life of me i cant remember my a-level maths!
    Any help is very much appreciated!
    Ben
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  2. #2
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    Start by taking ln of both sides. Then you can use a property of logarithms to bring x down from its perch.

    ln(30000)=o.1x*ln(60000)+0.9x*ln(10000)

    Factor out x, and bring some terms together.

    ln(30000)/(0.1*ln(60000)+0.9*ln(10000))=2x

    x=ln(30000)/2*(o.1*ln(60000)+0.9*ln(10000))

    This can further be reduced, but I'll leave that up to you.


    Best,


    -F
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  3. #3
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    Thank you! Just still very confused as to what to do next, I had got to that point but couldn't get it any further, sorry i should have explained in the question.!
    Ben
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  4. #4
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    Quote Originally Posted by FlacidCelery View Post
    Start by taking ln of both sides. <<< better not!
    Then you can use a property of logarithms to bring x down from its perch.

    ln(30000)=o.1x*ln(60000) + 0.9x*ln(10000)
    The +-sign idicates a product in the original equation which isn't there.

    Quote Originally Posted by bojker26 View Post
    solve x, I have really struggled with this.

    30,000=0.1(60,000)^x+0.9(10,000)^x
    For the life of me i cant remember my a-level maths!
    Any help is very much appreciated!
    Ben
    The only way I see to solve the equation is using an iterative method like Newton:

    Construct a function f:

    f(x)=60,000^x+9*10,000^x-300,000

    and calculate the zeros of this function.

    x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}

    I used x_0 = 1 and after few steps got a result of x \approx 1.069634491
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  5. #5
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    Quote Originally Posted by FlacidCelery View Post
    Start by taking ln of both sides. Then you can use a property of logarithms to bring x down from its perch.

    ln(30000)=o.1x*ln(60000)+0.9x*ln(10000)
    This is incorrect. log(ax+ by) is NOT alog(x)+ blog(y) so log(60000^{.1x}+ 10000^{.9x}\ne .1xln(60000)+ .9xln(10000)

    Earboth's numerical method is probably the best you can do.

    Factor out x, and bring some terms together.

    ln(30000)/(0.1*ln(60000)+0.9*ln(10000))=2x

    x=ln(30000)/2*(o.1*ln(60000)+0.9*ln(10000))

    This can further be reduced, but I'll leave that up to you.


    Best,


    -F
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