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  • The sixth term in the expansion of ( 1/x^8/3 + x^2 log x )^8 (where base is 10) is 5600. the number of values of x is

The sixth term in the expansion of ( 1/x^8/3 + x^2 log x )^8 (where base is 10) is 5600. the number of values of x is

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Solution:   We have ,   

                      (frac1x^frac83+x^2log_10x)^8

             Rightarrow         ^8c_5cdot frac1x^8cdot x^10(log_10x)^5=5600

            Rightarrow                         (log_10x)^5=frac560056x^2=(frac10x)^2

            Rightarrow                                  x=10.   

Posted by

Deependra Verma

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