Find the no. Of solution of equation   x2=2x.Please reply and send solutions to mobie number   +919587748173.Thank u.

palak singhal
31 Points
8 years ago
Arun Kumar IIT Delhi
8 years ago
Hi
$\\ x^2=2^x \\ seeing the RHS, LHS should be products of 2 only \\so 2^n \\2^{2n}=2^{2^n} \\=>2n=2^n \\=>n=1,2$
Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty
mycroft holmes
272 Points
8 years ago
If x>0, then consider the function, f(x) = x1/x. This is monotonic in the intervals (0, e) and (e, infty)

Thus only two solutions exist to x1/x =2 1/2 i.e. x =2 and x=4

If x<0, then f(x) = 2x – xmonotonically increasing and continuous. Also, f(0)>0 and f(-1)<0 and so one more solution may be found in this interval.

Thus three distinct solutions exist