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Math Problems
Algebra 2
Change of base formula
Evaluate the expression
7
2
x
2
−
2
\frac{7^{2}}{x^{2}-2}
x
2
−
2
7
2
for
x
=
3
x=3
x
=
3
.
\newline
□
\square
□
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NGETAHUAN KUANTITATIF KUANTI
24
24
24
−
12
-12
−
12
\newline
eskripsi Soal
\newline
Nilai bentuk
2
×
(
1
2
3
×
3
−
6
)
1
3
=
2 \times\left(\frac{1}{2^{3} \times 3^{-6}}\right)^{\frac{1}{3}}=
2
×
(
2
3
×
3
−
6
1
)
3
1
=
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aembeli jeruk dan mangga Gerobag buahnya hanya memuat
80
k
g
80 \mathrm{~kg}
80
kg
sedangkan harga
1
k
g
1 \mathrm{~kg}
1
kg
kan dibeli berturut - turut adalah
×
k
g
\times \mathrm{kg}
×
kg
dan
y
k
g
y \mathrm{~kg}
y
kg
maka model matematika permasalahar ersebut adalah.
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Grade HS Cla Rational numbers a
\newline
Name:
\newline
1
1
1
. Shade in the boxes of two numbers whose sum, when added, would be irrational.
\newline
\begin{tabular}{|c|c|c|c|c|}
\newline
\hline \begin{tabular}{c}
\newline
ratival \\
\newline
2
16
2 \sqrt{16}
2
16
\\
\newline
6
6
6
\newline
\end{tabular} & \begin{tabular}{c}
\newline
rationat \\
\newline
2
2
2
\newline
\end{tabular} & \begin{tabular}{c}
\newline
1
1
1
\\
\newline
5
5
5
\newline
\end{tabular} &
4
3
\frac{4}{3}
3
4
&
4
10
4 \sqrt{10}
4
10
\\
\newline
\hline
\newline
\end{tabular}
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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
)
(
3
x
−
5
)
=
2
\log _{(x+7)}(3 x-5)=2
lo
g
(
x
+
7
)
(
3
x
−
5
)
=
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
4
x
(
2
x
−
7
)
=
3
x
−
5
\log _{4 x}(2 x-7)=3 x-5
lo
g
4
x
(
2
x
−
7
)
=
3
x
−
5
\newline
Answer:
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LACER
\newline
MathML Display - This is not a re rification purposes only.
\newline
Instructions
\newline
Choose the best answer. If neces:
\newline
x
9
1049
−
1264
+
x
5
1849
−
x
7
11549
=
\frac{x^{9}}{1049}-1264+\frac{x^{5}}{1849}-\frac{x^{7}}{11549}=
1049
x
9
−
1264
+
1849
x
5
−
11549
x
7
=
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Rewrite the expression as a product of four linear factors:
\newline
(
x
2
+
5
x
)
2
+
10
(
x
2
+
5
x
)
+
24
\left(x^{2}+5 x\right)^{2}+10\left(x^{2}+5 x\right)+24
(
x
2
+
5
x
)
2
+
10
(
x
2
+
5
x
)
+
24
\newline
Answer:
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Simplify
e
3
ln
4
+
2
e^{3 \ln 4+2}
e
3
l
n
4
+
2
and write without any logarithms.
\newline
Answer:
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Write the expression
ln
3
−
ln
5
\ln 3-\ln 5
ln
3
−
ln
5
as a single logarithm in simplest form without any negative exponents.
\newline
Answer:
ln
(
□
)
\ln (\square)
ln
(
□
)
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We want to solve the following equation.
\newline
3
x
+
3
=
e
x
\sqrt{3 x+3}=e^{x}
3
x
+
3
=
e
x
\newline
One of the solutions is
x
≈
−
1
x \approx-1
x
≈
−
1
.
\newline
Find the other solution.
\newline
Hint: Use a graphing calculator.
\newline
Round your answer to the nearest tenth.
\newline
x
≈
x \approx
x
≈
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Consider the equation
\newline
7
⋅
e
0.1
t
=
47
.
7 \cdot e^{0.1 t}=47 \text {. }
7
⋅
e
0.1
t
=
47
.
\newline
Solve the equation for
t
t
t
. Express the solution as a logarithm in base
e
e
e
.
\newline
t
=
□
t=\square
t
=
□
\newline
Approximate the value of
t
t
t
. Round your answer to the nearest thousandth.
\newline
t
≈
t \approx
t
≈
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Consider the equation
\newline
2
⋅
e
−
6
w
=
95
.
2 \cdot e^{-6 w}=95 \text {. }
2
⋅
e
−
6
w
=
95
.
\newline
Solve the equation for
w
w
w
. Express the solution as a logarithm in basee.
\newline
w
=
w=
w
=
\newline
Approximate the value of
w
w
w
. Round your answer to the nearest thousandth.
\newline
w
≈
w \approx
w
≈
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Consider the equation
\newline
11
⋅
1
0
5
t
=
20
.
11 \cdot 10^{5 t}=20 \text {. }
11
⋅
1
0
5
t
=
20
.
\newline
Solve the equation for
t
t
t
. Express the solution as a logarithm in base
10
10
10
.
\newline
t
=
t=
t
=
\newline
Approximate the value of
t
t
t
. Round your answer to the nearest thousandth.
\newline
t
≈
t \approx
t
≈
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Consider the equation
\newline
0.5
⋅
e
4
z
=
13
.
0.5 \cdot e^{4 z}=13 \text {. }
0.5
⋅
e
4
z
=
13
.
\newline
Solve the equation for
z
z
z
. Express the solution as a logarithm in basee.
\newline
z
=
z=
z
=
\newline
Approximate the value of
z
z
z
. Round your answer to the nearest thousandth.
\newline
z
≈
z \approx
z
≈
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Consider the equation
\newline
9
⋅
e
2
z
=
54
.
9 \cdot e^{2 z}=54 \text {. }
9
⋅
e
2
z
=
54
.
\newline
Solve the equation for
z
z
z
. Express the solution as a logarithm in basee.
\newline
z
=
z=
z
=
\newline
Approximate the value of
z
z
z
. Round your answer to the nearest thousandth.
\newline
z
≈
z \approx
z
≈
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Consider the equation
\newline
4
⋅
1
0
−
3
x
=
18
4 \cdot 10^{-3 x}=18
4
⋅
1
0
−
3
x
=
18
\newline
Solve the equation for
x
x
x
. Express the solution as a logarithm in base
10
10
10
.
\newline
x
=
x=
x
=
\newline
Approximate the value of
x
x
x
. Round your answer to the nearest thousandth.
\newline
x
≈
x \approx
x
≈
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Consider the equation
\newline
−
5
⋅
e
10
t
=
−
30
.
-5 \cdot e^{10 t}=-30 \text {. }
−
5
⋅
e
10
t
=
−
30
.
\newline
Solve the equation for
t
t
t
. Express the solution as a logarithm in base
e
e
e
.
\newline
t
=
□
t=\square
t
=
□
\newline
Approximate the value of
t
t
t
. Round your answer to the nearest thousandth.
\newline
t
≈
t \approx
t
≈
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Consider the equation
\newline
6
⋅
e
0.25
t
=
9
.
6 \cdot e^{0.25 t}=9 \text {. }
6
⋅
e
0.25
t
=
9
.
\newline
Solve the equation for
t
t
t
. Express the solution as a logarithm in base
e
e
e
.
\newline
t
=
□
t=\square
t
=
□
\newline
Approximate the value of
t
t
t
. Round your answer to the nearest thousandth.
\newline
t
≈
t \approx
t
≈
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Consider the equation
\newline
4
⋅
e
2.7
x
=
33
.
4 \cdot e^{2.7 x}=33 \text {. }
4
⋅
e
2.7
x
=
33
.
\newline
Solve the equation for
x
x
x
. Express the solution as a logarithm in base
e
e
e
.
\newline
x
=
x=
x
=
\newline
Approximate the value of
x
x
x
. Round your answer to the nearest thousandth.
\newline
x
≈
x \approx
x
≈
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Consider the equation
\newline
5
⋅
e
−
7
x
=
12
.
5 \cdot e^{-7 x}=12 \text {. }
5
⋅
e
−
7
x
=
12
.
\newline
Solve the equation for
x
x
x
. Express the solution as a logarithm in basee.
\newline
x
=
x=
x
=
\newline
Approximate the value of
x
x
x
. Round your answer to the nearest thousandth.
\newline
x
≈
x \approx
x
≈
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Consider the equation
\newline
−
2
⋅
1
0
4
x
=
−
300
.
-2 \cdot 10^{4 x}=-300 \text {. }
−
2
⋅
1
0
4
x
=
−
300
.
\newline
Solve the equation for
x
x
x
. Express the solution as a logarithm in base
10
10
10
.
\newline
x
=
x=
x
=
\newline
Approximate the value of
x
x
x
. Round your answer to the nearest thousandth.
\newline
x
≈
x \approx
x
≈
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Consider the equation
\newline
−
3
⋅
1
0
2
t
=
−
28
.
-3 \cdot 10^{2 t}=-28 \text {. }
−
3
⋅
1
0
2
t
=
−
28
.
\newline
Solve the equation for
t
t
t
. Express the solution as a logarithm in base
10
10
10
.
\newline
t
=
t=
t
=
\newline
Approximate the value of
t
t
t
. Round your answer to the nearest thousandth.
\newline
t
≈
t \approx
t
≈
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Rewrite as a quotient of two common logarithms. Write your answer in simplest form.
\newline
log
3
33
=
\log_3 33 =
lo
g
3
33
=
______
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