Rewrite the expression
5
log
x
−
4
log
(
x
2
+
1
)
+
4
log
(
x
−
1
)
\displaystyle {5}{\log{{x}}}-{4}{\log{{\left({x}^{{2}}+{1}\right)}}}+{4}{\log{{\left({x}-{1}\right)}}}
5
lo
g
x
−
4
lo
g
(
x
2
+
1
)
+
4
lo
g
(
x
−
1
)
as a single logarithm
log
A
\displaystyle {\log{{A}}}
lo
g
A
. Then the function
A
=
\displaystyle {A}=
A
=
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