if 1,log 3(3x-2), 2log9(3x-8/3)are in the AP, then the value of x can be
Answers
Answer:
We have an AP: 1, log9(31−x+2), log3(4.3x−1) We have to find value of x.
d=a2−a1
d=log(31−x+2)9−1
d=log(31−x+2)123−log33
d=log(31−x+2)1233
Also
d=a3−a2
d=log(4.3x−1)3−log(31−x+2)123
d=log4.3x−1(31−x+2)123
now
a2−a1=a3−a2
log(31−x+2)1233=log4.3x−1(31−x+2)123
log(31−x+2)1233−log4.3x−1(31−x+2)123=0
log(31−x+2)123.(31−x+2)124.3x−13=0
log31−x+23(4.3x−1)3=0
so
31−x+23.4.3x−3=1
31−x+2=3.4.3x−3
3.4.3x−31−x=5
Answer:
Explanation:
So, we given 1,log
3
(3
x
−2),2log
9
(3
x
−8
3
) are in A.P
Then 2log
3
(3
x
−2)=2log
9
(3
x
−8
3
)+1
Note that log
9
(3
x
−8
3
)=
2
1
log
3
(3
x
−8
3
)
So 2log
3
(3
x
−2)=
2
2
log
3
(3
x
−8
3
)+log
3
3
2log
3
(3
x
−2)=log
3
3(3
x
−8
3
)
Let 3
x
=y2log
3
(y−2)=log
3
y−8
3
⇒(y−2)
2
=log
3
3(y−8
3
)
⇒y
2
−7y+12=0
⇒y=3 or y=4
⇒3
x
=3 or 3
x
=4
⇒x=log
3
3
or x=log
3
4
⇒x=y or xlog
3
4