5 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.1. The lhs expression increases faster than y=x after a certain point.
2. For there to be only one solution the minimum of the rhs must touch the y=x line.
3. Uske baad note that the minimum of the expression inside the log occirs when 3^x=k so putting that in, log_3(k^2-2k^2+4k-1)=log_3(k)
Solving this you get -k^2+3k-1=0 or k^2-3k+1=0
Check if the roots are in the domain and done.
Hello sir! Thanks for replying..This is the solution for the question. Why didn't we take the case where one root is positive and the other strictly negative(not non negative)? Are there only 3 k values as given there??
