Algebra+2+Quarter+3+Review+Assignment
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Apr 25, 2024
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Algebra 2 Third Quarter Review Assignment
Each problem below is similar to the questions on the 3
rd
Quarter Practice Exam.
You must show your work to receive credit! Question 1)
Identify the inverse of the function.
f
(
x
)
=
3
x
+
2
4
1.
Replace �
(
�
)
f
(
x
) with �
y
: �
=3
�
+24
y
=43
x
+2
.
2.
Swap �
x
and �
y
: �
=3
�
+24
x
=43
y
+2
.
3.
Solve for �
y
.
Starting from step 2:
�
=3
�
+24
x
=43
y
+2
Multiply both sides by 4 to get rid of the denominator:
4
�
=3
�
+24
x
=3
y
+2
Subtract 2 from both sides:
4
�
−2=3
�
4
x
−2=3
y
Finally, divide both sides by 3 to solve for �
y
:
�
=4
�
−23
y
=34
x
−2
So, the inverse function of �
(
�
)=3
�
+24
f
(
x
)=43
x
+2
is:
�
−1(
�
)=4
�
−23
f
−1(
x
)=34
x
−2
Question 2)
Solve the equation.
27
x
+
5
=
9
2
x
−
4
To solve the equation 27
�
+5=92
�
−427
x
+5
=9
2
x
−4
, notice that both 27 and 9 can be expressed as powers
of 3 since 27=3327=3
3
and 9=329=3
2
. Rewriting the equation using this insight:
33(
�
+5)=32(2
�
−4)3
3(
x
+5)
=3
2(2
x
−4)
Simplifying each side:
33
�
+15=34
�
−83
3
x
+15
=3
4
x
−8
Since the bases are the same and the equation is set, the exponents must be equal:
3
�
+15=4
�
−83
x
+15=4
x
−8
Now, solve for �
x
:
15+8=4
�
−3
�
15+8=4
x
−3
x
23=
�
23=
x
Therefore, the solution to the equation 27
�
+5=92
�
−427
x
+5
=9
2
x
−4
is �
=23
x
=23
.
Question 3)
Solve the equation.
lo g
3
(
4
x
−
2
)
−
lo g
3
2
=
2
1.
Combine the logarithmic expressions using the property that log
��
−log
��
=log
�
(
��
)log
bA
−log
b
B
=log
b
(
BA
):
log3(4
�
−22)=2log3(24
x
−2)=2
2.
Simplify the fraction inside the logarithm:
log3(2
�
−1)=2log3(2
x
−1)=2
3.
To get rid of the logarithm, convert the equation into its exponential form. This step is based on the definition of logarithms: if log
��
=
�
log
bA
=
C
, then ��
=
�
bC
=
A
:
32=2
�
−132=2
x
−1
4.
Solve the resulting equation:
32=9=2
�
−132=9=2
x
−1
Adding 1 to both sides gives:
9+1=2
�
9+1=2
x
10=2
�
10=2
x
Divide both sides by 2 to solve for �
x
:
�
=102=5
x
=210=5
Therefore, the solution to the equation log3(4
�
−2)−log32=2log3(4
x
−2)−log32=2 is �
=5
x
=5.
Question 4)
Solve the equation.
x
−
21
x
=−
4
To solve the equation �
−21
�
=−4
x
−
x
21
=−4, it's convenient to start by getting rid of the fraction. First, let's rewrite the equation to see it clearly:
�
−21
�
=−4
x
−
x
21
=−4
Multiply every term by �
x
to eliminate the fraction:
�
2−21=−4
�
x
2−21=−4
x
Rearrange the equation to set it to 0, moving all terms to one side (ideal for solving quadratic equations):
�
2+4
�
−21=0
x
2+4
x
−21=0
Now, we have a quadratic equation in the form of ��
2+
��
+
�
=0
ax
2+
bx
+
c
=0. To solve it, you can either factorize it (if possible), complete the square, or use the quadratic formula. The quadratic formula is:
�
=−
�
±
�
2−4
��
2
�
x
=2
a
−
b
±
b
2−4
ac
For our equation, �
=1
a
=1, �
=4
b
=4, and �
=−21
c
=−21. Plugging these into the quadratic formula gives:
�
=−4±42−4(1)(−21)2(1)
x
=2(1)−4±42−4(1)(−21)
�
=−4±16+842
x
=2−4±16+84
�
=−4±1002
x
=2−4±100
�
=−4±102
x
=2−4±10
This gives us two possible solutions for �
x
:
1.
�
=−4+102=62=3
x
=2−4+10
=26
=3
2.
�
=−4−102=−142=−7
x
=2−4−10
=2−14
=−7
Therefore, the solutions to the equation �
−21
�
=−4
x
−
x
21
=−4 are �
=3
x
=3 and �
=−7
x
=−7.
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