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Math Problems
Precalculus
Convert between exponential and logarithmic form
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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log
(
P
1
)
=
log
(
17
⋅
(
P
2
)
2
)
\log \left(P_{1}\right)=\log \left(17 \cdot\left(P_{2}\right)^{2}\right)
lo
g
(
P
1
)
=
lo
g
(
17
⋅
(
P
2
)
2
)
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log
(
P
1
)
=
log
(
17
x
(
P
2
)
2
)
\log \left(P_{1}\right)=\log \left(17 x\left(P_{2}\right)^{2}\right)
lo
g
(
P
1
)
=
lo
g
(
17
x
(
P
2
)
2
)
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log
x
(
1
81
)
=
−
3
\log_{x}(\frac{1}{81}) = -3
lo
g
x
(
81
1
)
=
−
3
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Find the area of the shaded region. Round to the nearest tenth if necessary.
\newline
log
=
\log=
lo
g
=
\newline
v
e
2
ve^{2}
v
e
2
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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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Pythagoras theorem
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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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Find the equation of the exponential function represented by the table below:
\newline
\begin{tabular}{|c|c|}
\newline
\hline
x
x
x
&
y
y
y
\\
\newline
\hline
0
0
0
&
5
5
5
\\
\newline
\hline
1
1
1
&
2
2
2
.
5
5
5
\\
\newline
\hline
2
2
2
&
1
1
1
.
25
25
25
\\
\newline
\hline
3
3
3
&
0
0
0
.
625
625
625
\\
\newline
\hline
\newline
\end{tabular}
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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
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
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
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
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
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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pythagoras theorem
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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
(
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
(
x
−
7
)
(
6
)
=
2
x
\log _{(x-7)}(6)=2 x
lo
g
(
x
−
7
)
(
6
)
=
2
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
x
(
5
x
)
=
2
\log _{5 x}(5 x)=2
lo
g
5
x
(
5
x
)
=
2
\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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Write in exponential notation:
\newline
(
(
n
4
)
5
)
2
\left(\left(n^{4}\right)^{5}\right)^{2}
(
(
n
4
)
5
)
2
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ext{ exttt{log}}_{
2
2
2
}(
4
4
4
)=
2
2
2
\rightarrow
2
2
2
^{
2
2
2
}=
4
4
4
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Write in exponential notation:
\newline
(
3
m
)
2
(3m)^2
(
3
m
)
2
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Solve the equation.
\newline
log
(
x
+
5
)
=
log
(
4
x
+
2
)
\log(x+5)=\log(4x+2)
lo
g
(
x
+
5
)
=
lo
g
(
4
x
+
2
)
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Exponential and Logarithmic Functions
\newline
Using a graphing calculator to solve an exponential or logarithmic...
\newline
Use the ALEKS graphing calculator to solve the equation.
\newline
e
2
−
3
x
=
5
−
3
x
e^{2-3x}=5-3x
e
2
−
3
x
=
5
−
3
x
\newline
Round to the nearest hundredth.
\newline
If there is more than one solution, separate them with commas.
\newline
x
=
□
x=\square
x
=
□
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28
28
28
.
log
(
x
+
1
)
=
log
x
+
1
\log (x+1)=\log x+1
lo
g
(
x
+
1
)
=
lo
g
x
+
1
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Write in standard form:
\newline
6
×
1
0
5
6 \times 10^{5}
6
×
1
0
5
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Write an exponential function in the form
y
=
a
b
x
y=a b^{x}
y
=
a
b
x
that goes through the points
(
0
,
14
)
(0,14)
(
0
,
14
)
and
(
4
,
1134
)
(4,1134)
(
4
,
1134
)
.
\newline
Answer:
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Write an exponential function in the form
y
=
a
b
x
y=a b^{x}
y
=
a
b
x
that goes through the points
(
0
,
15
)
(0,15)
(
0
,
15
)
and
(
4
,
1215
)
(4,1215)
(
4
,
1215
)
.
\newline
Answer:
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Find the derivative of the following function.
\newline
y
=
log
5
(
6
x
6
−
2
x
5
)
y=\log _{5}\left(6 x^{6}-2 x^{5}\right)
y
=
lo
g
5
(
6
x
6
−
2
x
5
)
\newline
Answer:
y
′
=
y^{\prime}=
y
′
=
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