If the value of the third term in the expansion of x+xlog10x5is 106 , then x may have value(s)
1, 10
10,10−5/2
10,10−3/2
102,10−3/2
T3=5C2x5−2xlog10x2=106⇒x2log10x+3=105
⇒t(2t+3)=5 where t=log10x⇒t=−5/2,1