If the last term in the binomial expansion of 21/3−12n is 135/3log3⁡8, the the 5th term from  the begging is

Last term of  21/3−12n is Tn+1=nCn21/3n−n−12n=nCn(−1)n12n/2=(−1)n2n/2Also, we have135/3log3⁡8=3−(5/3)log3⁡23=2−5 Thus, (−1)n2n/2=2−5⇒ (−1)n2n/2=(−1)1025⇒ n2=5⇒n=10Now T5=T4+1=10C421/310−4−124=10!4!6!21/36(−1)42−1/24=210(2)2(1)2−2=210