The value of x, for which the 6th term in the expansion of 2log29x−1+7+1215log23x−1+17is 84 is
4
3
2
1
E=9x−1+7+13x−1+11/57
Given T6=84
⇒7C59x−1+77−513x−1+11/55=84or 7C59x−1+7⋅13x−1+1=84or 9x−1+7=43x−1+1or 32x−12⋅3x+27=0or 3x−33x−9=0or 3x=3,9x=1,2