Question 303232
{{{log((2x^2))=(log((2x)))^2}}}
I used Newton's method to find the roots of 
{{{f(x)=log((2x^2))-(log((2x)))^2}}}
{{{df/dx=2/x-(2*log(2x))/x}}}
and I started with a value of x=0.2.
A quick check in EXCEL showed that the zero was between x=0.7 and x=0.8. 
After 20 iterations, I got
{{{x=0.7292}}}
Quick check to verify,
{{{log((2(0.7292)^2))=log((2(0.7292)))^2}}}
{{{log((2(0.7292)^2))=log((2(0.7292)))^2}}}
{{{log(1.0634)=(log(1.4584))^2}}}
{{{0.026697=(.16387)^2}}}
{{{0.026697=0.026856}}}