Values of x for which the sixth term of the expansion of
E=(3log3 √9|x−2|+7(15)log7 [(4).3|x−2|−9])7 is 567, are
1
3
Put y=3log3√p|x−2|⇒log3 y=log3 √9|x−2|
⇒y=√9|x−2|=3|x−2|
Next, put z=(715log7 [(4).3x−2−9])
⇒ log7 z=15 log7[(4)3|x−2|−9]=log7[(4)3|x−2|−9]15⇒ z=[(4)3|x−2|−9]15
Now, E=(y+z)7 and 6th term is given by
t6=7C5 y7−5 z5=21(3|x−2|)2{(4)3|x−2|−9}⇒ 567=21{32|x−2|}{(4)3|x−2|−9}⇒ 27=(4)33|x−2|−(9)(32|x−2|)⇒ 27=(4)33|x−2|−(9)(32|x−2|)⇒ 4u3−9u2−27=0 where u=3|x−2|
Note that u=3 satisfies this equation
∴ 3|x−2|=3⇒|x−2|=1⇒x−2=±1x=2±1=3 or 1