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Mathematics
Mathematics
Problem
The equation
2
=
log
x
+
1
(
y
+
1
)
2 = \log_{x+1}(y + 1)
2
=
lo
g
x
+
1
(
y
+
1
)
can be written as: A.
y
=
2
log
x
+
1
−
2
y = \frac{2}{\log_{x+1}} - 2
y
=
l
o
g
x
+
1
2
−
2
B.
y
=
(
x
+
1
)
2
−
1
y = (x + 1)^2 - 1
y
=
(
x
+
1
)
2
−
1
C.
y
=
2
(
x
+
1
)
−
1
y = 2(x + 1) - 1
y
=
2
(
x
+
1
)
−
1
D.
y
=
log
x
+
1
2
−
1
y = \log_{x+1} 2 - 1
y
=
lo
g
x
+
1
2
−
1
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