Given that 5^2x-6×5^x+5=0
Put 5^x=t - - - - - - - - - -- -- - - -:- (1), so we get,
t²-6t+5=0 , which is quadratic equation in t.
=> t²-t-5t+5=0 => (t²-t)-(5t-5)=0
=> t(t-1)-5(t-1)=0 => (t-1)(t-5)=0
=> Either t-1=0 or t-5=0
=> t=1 or t=5
=> t=1, 5
Case1:- When t=1, then 5^x=1 [By using (1)]
=> 5^x=5^0 => x=0
Case2:- When t=5
=> 5^x=5 => 5^x=5^1
=> x=1
From case1 and case2, we get, x=0,1 Ans.





