3log325x−1+7+3(−81)log3(5x−1+1)10
=25x−1+7+(5x−1+1)8−110, using anlogax=xn
Using the formula of the general term in the Binomial theorem for non negative integral index, i.e., Tr+1=Crnan−rbr in (a+b)n, we get, T9=T8+1=C810(25(x−1)+7)×(5(x−1)+1)−1=180
⇒5(x−1)+125x−1+7=4
⇒t+1t2+7=4, taking t=5x−1
⇒t2+7=4(t+1)
⇒t2−4t+3=0
⇒(t−3)(t−1)=0
⇒t=1,3⇒5x−1=1⇒5x−1=50
⇒x−1=0 (one of the possible value).
⇒x=1