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Hard maths equation!

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!
Reply 1
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
Reply 2
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 ?
Yes this cannot be solved algebraically ^^.

Was trying for the last ten minutes :frown:
Reply 4
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.
Reply 5

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
Reply 6
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.
Reply 7
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 :smile:

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