First slide
Binomial theorem for positive integral Index
Question

If the value of the third term in the expansion of x+xlog10x5is 106  , then x may have value(s) 

Moderate
Solution

T3=5C2x52xlog10x2=106x2log10x+3=105

t(2t+3)=5 where t=log10xt=5/2,1

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