log8^x+log4^x+log2^x=11 find x
Answers
Answered by
0
Answer:
As
log
a
b
=
log
b
log
a
, we can write
log
2
x
+
log
4
x
+
log
8
x
=
11
as
log
x
log
2
+
log
x
log
4
+
log
x
log
8
=
11
or
log
x
log
2
+
log
x
2
log
2
+
log
x
3
log
2
=
11
or
log
x
log
2
(
1
+
1
2
+
1
3
)
=
11
or
log
x
log
2
(
6
+
3
+
2
6
)
=
11
or
log
x
log
2
(
11
6
)
=
11
or
log
x
log
2
=
11
×
6
11
=
6
i.e.
log
x
=
6
log
2
=
log
2
6
and
x
=
2
6
=
64
Step-by-step explanation:
hope this helps you please mark as brainlest answer
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