solve for x in logx^2 = x/25
Welcome to Open Study Please post in the the proper section; this is a math question in the chemistry section. thanks.
Hmm it seems like you might want to get rid of the logarithm function first, no? (that's my first instinct) If that's log base 10, then you need to multiply 10^( ) to both sides. If that's the natural log, then you need to multiply e^( ) to both sides. Assuming it's base 10... \[\log_{10}x^2=\frac{x}{25}\] \[\huge 10^{log_{10}x^2}=10^{x/25}\] \[x^2=10^{x/25}\] \[x=\pm\sqrt{10^{x/25}}\] Where to go from there I'm not sure...
I would start like\[\log(x^2)=\frac x{25}\]\[2\log x=\frac x{25}\]\[\log x=\frac x{50}\]\[x=10^{x/50}\]\[10^{x/50}-x=0\]then use ralphson-newton approximation
hello Turing Test may go on please with ralphson-newton approximation
hey turing may you go on
oh that's a whole to-do involving calculus do you know calculus?
hello Turing may progress with the question then
do you know calculus?
not that much
may you try going on with the question
not if you don't know calculus; it's quite a long process if you just want the answer I suggest you just plug this into wolfram http://www.wolframalpha.com/input/?i=&t=crmtb01
graph the two functions and see where they meet x= 1.0495191898... or x=100.
how can you graph it Pax
http://lmgtfy.com/?q=y%3Dlog(x%5E2)%2C+y%3Dx%2F25 other than \(\large x=100 \), you also get \(\large x\approx-0.95689 \) and \(\large x\approx 1.04952\) http://www.wolframalpha.com/input/?i=log_10%28x^2%29%3Dx%2F25
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