SOLUTION: i need help with this prob please show work so i know what to do log 2x^2= (log 2x)^2

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: i need help with this prob please show work so i know what to do log 2x^2= (log 2x)^2      Log On


   

Question 303232: i need help with this prob please show work so i know what to do
log 2x^2= (log 2x)^2
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
log%28%282x%5E2%29%29=%28log%28%282x%29%29%29%5E2
I used Newton's method to find the roots of
f%28x%29=log%28%282x%5E2%29%29-%28log%28%282x%29%29%29%5E2
df%2Fdx=2%2Fx-%282%2Alog%282x%29%29%2Fx
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%28%282%280.7292%29%5E2%29%29=log%28%282%280.7292%29%29%29%5E2
log%28%282%280.7292%29%5E2%29%29=log%28%282%280.7292%29%29%29%5E2
log%281.0634%29=%28log%281.4584%29%29%5E2
0.026697=%28.16387%29%5E2
0.026697=0.026856