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Byju's Answer
Standard XII
Mathematics
Continuity in an Interval
Evaluate: l...
Question
Evaluate:
log
3
(
1
+
1
3
)
+
log
3
(
1
+
1
4
)
+
.
.
.
.
.
.
.
+
log
3
(
1
+
1
80
)
Open in App
Solution
l
o
g
3
(
1
+
1
3
)
+
l
o
g
3
(
1
+
1
4
)
+
l
o
g
3
(
1
+
1
5
)
.
.
.
.
.
+
l
o
g
3
(
1
+
1
79
)
+
l
o
g
3
(
1
+
1
80
)
⇒
l
o
g
3
(
4
3
)
+
l
o
g
3
(
5
4
)
+
l
o
g
3
(
6
5
)
.
.
.
.
.
+
l
o
g
3
(
80
79
)
+
l
o
g
3
(
81
80
)
⇒
l
o
g
3
(
4
3
×
5
4
×
6
5
.
.
.
.
80
79
×
81
80
)
⇒
l
o
g
3
(
1
3
×
81
1
)
⇒
l
o
g
3
(
27
)
⇒
l
o
g
3
(
3
3
)
⇒
3
l
o
g
3
(
3
)
=
3
×
1
=
3
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0
Similar questions
Q.
log
3
(
1
+
1
3
)
+
log
3
(
1
+
1
4
)
+
log
3
(
1
+
1
5
)
+
.
.
.
+
log
3
(
1
+
1
242
)
when simplified has the value equal to
Q.
1
log
3
e
+
1
log
3
e
2
+
1
log
3
e
4
+
.
.
.
.
∞
is
Q.
The value of
1
log
3
e
+
1
log
3
e
2
+
1
log
3
e
4
+
.
.
.
up to infinite terms is
Q.
The value of
x
satisfying
log
3
4
−
2
log
3
√
3
x
+
1
=
1
−
log
3
(
5
x
−
2
)
is:
Q.
The value of
x
satisfying the equation
log
3
(
5
x
−
2
)
−
2
log
3
√
3
x
+
1
=
1
−
log
3
4
is
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