First, remember that:
log
a
(
b
⋅
c
)
=
log
a
(
b
)
+
log
a
(
c
)
Therefore, we can say that:
log
(
4
x
)
+
log
(
3
x
+
1
)
=
log
(
4
x
⋅
(
3
x
+
1
)
)
⇒
log
(
12
x
2
+
4
x
)
We now rewrite the equation:
log
(
12
x
2
+
4
x
)
=
3
Now remember that:
log
(
b
)
=
log
10
(
b
)
And that if
log
a
(
b
)
=
c
, then
b
=
a
c
Therefore, we can now simplify the equation.
log
(
12
x
2
+
4
x
)
=
3
=>
log
10
(
12
x
2
+
4
x
)
=
3
=>
(
12
x
2
+
4
x
)
=
10
3
=>
12
x
2
+
4
x
=
1000
=>
12
x
2
+
4
x
−
1000
=
0
Let's use the quadratic formula
−
b
±
√
b
2
−
4
(
a
)
(
c
)
2
(
a
)
We now know that:
x
=
−
4
±
√
4
2
−
4
(
12
)
(
−
1000
)
2
(
12
)
We can now solve for
x
.
=>
x
=
−
4
±
√
48016
24
=>
x
=
−
4
±
4
√
3001
24
=>
x
=
−
1
±
√
3001
6
=>
x
≈
8.964
,
9.197
Now, we see that when
x
<
0
, whatever solution we have will be extraneous.
Therefore,
x
≈
8.964