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Math Problems
Algebra 1
Power rule with rational exponents
Write the expression
1
3
ln
27
−
3
ln
5
\frac{1}{3} \ln 27-3 \ln 5
3
1
ln
27
−
3
ln
5
as a single logarithm in simplest form without any negative exponents.
\newline
Answer:
ln
(
□
)
\ln (\square)
ln
(
□
)
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Write the expression
−
ln
2
+
ln
5
-\ln 2+\ln 5
−
ln
2
+
ln
5
as a single logarithm in simplest form without any negative exponents.
\newline
Answer:
ln
(
□
)
\ln (\square)
ln
(
□
)
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Write the expression
1
3
ln
8
+
ln
3
\frac{1}{3} \ln 8+\ln 3
3
1
ln
8
+
ln
3
as a single logarithm in simplest form without any negative exponents.
\newline
Answer:
ln
(
□
)
\ln (\square)
ln
(
□
)
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g
(
x
)
=
2
x
−
9
g
′
(
x
)
=
?
\begin{array}{l} g(x)=\sqrt{2 x-9} \\ g^{\prime}(x)=? \end{array}
g
(
x
)
=
2
x
−
9
g
′
(
x
)
=
?
\newline
Choose
1
1
1
answer:
\newline
(A)
2
x
−
9
2
\frac{\sqrt{2 x-9}}{2}
2
2
x
−
9
\newline
(B)
1
2
x
−
9
\frac{1}{\sqrt{2 x-9}}
2
x
−
9
1
\newline
(C)
1
2
2
x
−
9
\frac{1}{2 \sqrt{2 x-9}}
2
2
x
−
9
1
\newline
(D)
1
x
\frac{1}{\sqrt{x}}
x
1
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Let
f
(
x
)
=
1
∣
x
∣
f(x)=\frac{1}{|x|}
f
(
x
)
=
∣
x
∣
1
.
\newline
Can we use the mean value theorem to say the equation
f
′
(
x
)
=
−
1
4
f^{\prime}(x)=-\frac{1}{4}
f
′
(
x
)
=
−
4
1
has a solution where
2
<
x
<
4
2<x<4
2
<
x
<
4
?
\newline
Choose
1
1
1
answer:
\newline
(A) No, since the function is not differentiable on that interval.
\newline
(B) No, since the average rate of change of
f
f
f
over the interval
2
≤
x
≤
4
2 \leq x \leq 4
2
≤
x
≤
4
isn't equal to
−
1
4
-\frac{1}{4}
−
4
1
.
\newline
(C) Yes, both conditions for using the mean value theorem have been met.
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Simplify. Assume all variables are positive.
\newline
(
2
r
1
3
)
8
(2r^{\frac{1}{3}})^8
(
2
r
3
1
)
8
\newline
\newline
Write your answer in the form
A
A
A
or
A
B
\frac{A}{B}
B
A
, where
A
A
A
and
B
B
B
are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.
\newline
______
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