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    Hi, I'm getting stuck on the equation 50e^x(10x)=20
    I can work it down to x + ln(x) = ln (1/25), which I'm pretty sure is correct.
    But I dont know how to work the ln bit out at all.

    Any help would be appreciated, thanks!
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    (Original post by Student10011)
    Hi, I'm getting stuck on the equation 50e^x(10x)=20
    I can work it down to x + ln(x) = ln (1/25), which I'm pretty sure is correct.
    But I dont know how to work the ln bit out at all.

    Any help would be appreciated, thanks!
    That cannot be solved algebraically, you would need to use Numerical Methods
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    (Original post by Student10011)
    Hi, I'm getting stuck on the equation 50e^x(10x)=20
    I can work it down to x + ln(x) = ln (1/25), which I'm pretty sure is correct.
    But I dont know how to work the ln bit out at all.

    Any help would be appreciated, thanks!
    do you mean 50e^(10x^2)=20 ?
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    Yes this cannot be solved algebraically ^^.

    Was trying for the last ten minutes
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    (Original post by alow)
    do you mean 50e^(10x^2)=20 ?
    No, just 10x, it can be solved, I have an estimate of around 0.925/25. It seems from what others say, it can't be done algebraically. Thanks though.
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    SOLVED IT! Did it myself. I made the equation an iterative formula, and after many = presses on the calc, it terminated at a finite decimal number of 0.03848966594!

    Thanks everyone
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    (Original post by Student10011)

    SOLVED IT! Did it myself. I made the equation an iterative formula, and after many = presses on the calc, it terminated at a finite decimal number of 0.03848966594!

    Thanks everyone
    Look up the Newton-Raphson method if you're interested in solving by numerical methods.
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    (Original post by Student10011)

    SOLVED IT! Did it myself. I made the equation an iterative formula, and after many = presses on the calc, it terminated at a finite decimal number of 0.03848966594!

    Thanks everyone
    Well done - discovering maths for yourself is great
 
 
 
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