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Precalculus
Convert between exponential and logarithmic form
Write the exponential equation in logarithmic form.
\newline
e
1
/
3
≈
1.3956
e^{1/3} \approx 1.3956
e
1/3
≈
1.3956
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Convert the exponential equation in logarithmic form.
\newline
e
−
5
≈
0.0067
e^-5 \approx 0.0067
e
−
5
≈
0.0067
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Convert the exponential equation in logarithmic form.
\newline
e
−
4
≈
0.0183
e^-4 \approx 0.0183
e
−
4
≈
0.0183
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Convert the exponential equation in logarithmic form.
\newline
e
−
2
≈
0.1353
e^-2 \approx 0.1353
e
−
2
≈
0.1353
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Convert the exponential equation in logarithmic form.
\newline
e
−
1
≈
0.3679
e^-1 \approx 0.3679
e
−
1
≈
0.3679
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Convert the exponential equation in logarithmic form.
\newline
e
−
3
≈
0.0498
e^-3 \approx 0.0498
e
−
3
≈
0.0498
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Write the exponential equation in logarithmic form.
62
5
3
4
=
125
\newline625^{\frac{3}{4}} = 125
62
5
4
3
=
125
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Write the exponential equation in logarithmic form.
1
6
5
4
=
32
\newline16^{\frac{5}{4}} = 32
1
6
4
5
=
32
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Write the exponential equation in logarithmic form.
2
7
2
3
=
9
\newline 27^{\frac{2}{3}} = 9
2
7
3
2
=
9
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Convert the exponential equation in logarithmic form.
\newline
8
−
2
=
1
64
8^{-2} = \frac{1}{64}
8
−
2
=
64
1
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Convert the exponential equation in logarithmic form.
\newline
3
−
3
=
1
27
3^{-3} = \frac{1}{27}
3
−
3
=
27
1
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What is the domain of this logarithmic function?
y
=
log
3
(
x
+
9
)
+
1
\newline y=\log_3(x+9)+1\newline
y
=
lo
g
3
(
x
+
9
)
+
1
Express the domain in inequality notation.
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What is the domain of this logarithmic function?
y
=
log
2
(
x
−
6
)
−
8
\newline y=\log_2(x-6)-8\newline
y
=
lo
g
2
(
x
−
6
)
−
8
Express the domain in inequality notation.
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What is the domain of this logarithmic function?
y
=
log
5
(
x
−
2
)
+
7
\newline y=\log_5(x-2)+7\newline
y
=
lo
g
5
(
x
−
2
)
+
7
Express the domain in inequality notation.
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What is the domain of this exponential function
?
?
?
\newline
y
=
(
2
7
)
x
+
9
y = \left(\frac{2}{7}\right)^{x+9}
y
=
(
7
2
)
x
+
9
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What is the domain of this exponential function
?
?
?
\newline
y
=
(
9
)
x
+
1
y = (9)^{x+1}
y
=
(
9
)
x
+
1
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What is the domain of this exponential function
?
?
?
\newline
y
=
(
3
2
)
x
+
6
y = \left(\frac{3}{2}\right)^{x+6}
y
=
(
2
3
)
x
+
6
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What is the domain of this exponential function
?
?
?
\newline
y
=
(
1
8
)
x
+
2
y = \left(\frac{1}{8}\right)^{x+2}
y
=
(
8
1
)
x
+
2
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Write the expression in exponential form
8
8
8
x
8
8
8
x
8
8
8
x
8
8
8
.
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Write the expression in exponential form
6
×
6
×
6
×
6
×
6
6 \times 6 \times 6 \times 6 \times 6
6
×
6
×
6
×
6
×
6
.
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Write the expression in exponential form
9
×
9
×
9
×
9
×
9
×
9
×
9
9 \times 9 \times 9 \times 9 \times 9 \times 9 \times 9
9
×
9
×
9
×
9
×
9
×
9
×
9
.
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Solve for
x
x
x
\newline
log
6
x
=
log
x
36
\log _{6} x=\log _{x} 36
lo
g
6
x
=
lo
g
x
36
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Simplify:
log
x
(
1
81
)
=
−
3
\log_{x}(\frac{1}{81}) = -3
lo
g
x
(
81
1
)
=
−
3
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Solve for a positive value of
x
x
x
.
\newline
log
2
(
x
)
=
6
\log _{2}(x)=6
lo
g
2
(
x
)
=
6
\newline
Answer:
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Solve for
k
k
k
.
\newline
log
10
19
k
=
19
log
10
k
\log_{10}19k = 19\log_{10}k
lo
g
10
19
k
=
19
lo
g
10
k
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log
4
(
x
+
3
)
=
2
\log _{4}(x+3)=2
lo
g
4
(
x
+
3
)
=
2
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If
log
7
5
=
p
\log_7{5}=p
lo
g
7
5
=
p
and
log
10
2
=
q
\log_{10}{2} = q
lo
g
10
2
=
q
, solve
log
7
2
\log_7{2}
lo
g
7
2
in terms of
p
p
p
and
q
q
q
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Consider the equation
9
⋅
e
2
z
=
54
9 \cdot e^{2 z}=54
9
⋅
e
2
z
=
54
.
\newline
Solve the equation for
z
z
z
. Express the solution as a logarithm in base-
e
e
e
.
\newline
z
=
□
z = \square
z
=
□
\newline
Approximate the value of
z
z
z
. Round your answer to the nearest thousandth.
\newline
z
≈
□
z \approx \square
z
≈
□
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Condense the logarithm
\newline
log
b
+
g
log
c
\log b+g \log c
lo
g
b
+
g
lo
g
c
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
8
log
c
−
4
log
d
8 \log c-4 \log d
8
lo
g
c
−
4
lo
g
d
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
log
a
+
z
log
g
\log a+z \log g
lo
g
a
+
z
lo
g
g
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
q
log
a
−
log
b
q \log a-\log b
q
lo
g
a
−
lo
g
b
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
log
d
+
2
log
q
\log d+2 \log q
lo
g
d
+
2
lo
g
q
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
5
log
d
+
log
x
5 \log d+\log x
5
lo
g
d
+
lo
g
x
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
g
log
a
+
log
d
g \log a+\log d
g
lo
g
a
+
lo
g
d
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
k
log
a
+
g
log
d
k \log a+g \log d
k
lo
g
a
+
g
lo
g
d
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
2
log
a
−
8
log
b
2 \log a-8 \log b
2
lo
g
a
−
8
lo
g
b
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
g
log
a
+
log
q
g \log a+\log q
g
lo
g
a
+
lo
g
q
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
log
d
−
7
log
k
\log d-7 \log k
lo
g
d
−
7
lo
g
k
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
3
log
a
+
log
c
3 \log a+\log c
3
lo
g
a
+
lo
g
c
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
y
log
d
−
log
x
y \log d-\log x
y
lo
g
d
−
lo
g
x
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
7
log
a
−
x
log
d
7 \log a-x \log d
7
lo
g
a
−
x
lo
g
d
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
k
log
c
−
log
d
k \log c-\log d
k
lo
g
c
−
lo
g
d
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
7
log
b
−
log
k
7 \log b-\log k
7
lo
g
b
−
lo
g
k
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Condense the logarithm
\newline
3
log
a
+
log
k
3 \log a+\log k
3
lo
g
a
+
lo
g
k
\newline
Answer:
log
(
□
)
\log (\square)
lo
g
(
□
)
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Write the log equation as an exponential equation. You do not need to solve for
x
\mathrm{x}
x
.
\newline
log
(
x
)
=
2
x
+
1
\log (x)=2 x+1
lo
g
(
x
)
=
2
x
+
1
\newline
Answer:
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Write the log equation as an exponential equation. You do not need to solve for
x
\mathrm{x}
x
.
\newline
log
2
x
(
4
x
+
2
)
=
2
x
+
3
\log _{2 x}(4 x+2)=2 x+3
lo
g
2
x
(
4
x
+
2
)
=
2
x
+
3
\newline
Answer:
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Write the log equation as an exponential equation. You do not need to solve for
x
\mathrm{x}
x
.
\newline
log
(
x
+
6
)
(
5
)
=
x
\log _{(x+6)}(5)=x
lo
g
(
x
+
6
)
(
5
)
=
x
\newline
Answer:
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Write the log equation as an exponential equation. You do not need to solve for
x
\mathrm{x}
x
.
\newline
log
(
x
+
4
)
(
5
x
)
=
x
\log _{(x+4)}(5 x)=x
lo
g
(
x
+
4
)
(
5
x
)
=
x
\newline
Answer:
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Write the log equation as an exponential equation. You do not need to solve for
x
\mathrm{x}
x
.
\newline
log
(
5
)
=
2
x
−
5
\log (5)=2 x-5
lo
g
(
5
)
=
2
x
−
5
\newline
Answer:
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