If log2, log4 and logx are in AP, then find the value of ‘x’.
Answers
Answer:
Answer is in the picture attached
Step-by-step explanation:
log
2
(5.2
x
+1),log
4
(2
1−x
+1)and1 are in AP.
Common difference d=a
2
−a
1
=a
3
−a
2
∴log
4
(2
1−x
+1)−log
2
(5.2
x
+1)=1−log
4
(2
1−x
+1)
⇒log
4
(2
1−x
+1)+log
4
(2
1−x
+1)=log
2
2+log
2
(5.2
x
+1)[log
a
a=1]
⇒2log
4
(2
1−x
+1)=log
2
2+log
2
(5.2
x
+1)
⇒
2
2
⋅log
2
(2
1−x
+1)=log
2
(2(5.2
x
+1))
[Since,log
b
n
a=nlog
b
a & log
c
a+log
c
b=log
c
(ab)]
∴2
1−x
+1=5.2
x+1
+2
⇒2⋅
2
x
1
=10⋅2
x
+1
Let 2
x
=y
∴
y
2
=10y+1
⇒10y
2
+y−2=0
⇒10y
2
+5y−4y−2=0
⇒5y(2y+1)−2(2y+1)=0
⇒(5y−2)(2y+1)=0
⇒y=
5
2
ory=
2
−1
⇒2
x
=
5
2
or2
x
=
2
−1
∴2
x
=
5
2
(Since exponent of any real number can never be a negative value).
Applying log
2
on both sides,
log
2
(2x)=log
2
(
5
2
)
⇒x=log
2
2−log
2
5
⇒x=1−log
2
5