# Mathematics 105 Practice Final Exam
**May, 2025**

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## Instructions:
- Answer the questions clearly.
- No calculators, books, or notes.
- Show all work.

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## Formulas:

$$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

$$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$$

$$\frac{y_2-y_1}{x_2-x_1}$$

$$y-y_1 = m(x-x_1)$$

$$P(x) = R(x) - C(x)$$

$$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

$$A = P\left(1 + \frac{r}{n}\right)^{nt}$$

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## Questions:

### Question 1: Distance Between Two Points
Find the distance between the points $(2, -4)$ and $(5, 1)$.
Express your answer in the form $A\sqrt{B}$, where perfect squares are removed from under the square root.

### Solution:

\begin{align}
d&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\
&=\sqrt{(5-2)^2+(1-(-4))^2} \\
&=\sqrt{3^2+(1+4)^2} \\
&=\sqrt{3^2+5^2} \\
&=\sqrt{9+25} \\
&=\sqrt{34}
\end{align}

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### Question 2: Numerical Evaluation
Evaluate the following:

1. $(193.76)(100)=$
2. $0.06763 \div 100=$
3. $-7.583 / 1000=$

### Solution:

1. $(193.76)(100)=19376$
2. $0.06763 \div 100=0.0006763$
3. $-7.583 / 1000=-0.007583$

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### Question 3: Midpoint of a Segment
Find the midpoint of the segment having endpoints $(2, -7)$ and $(5, 18)$.
Your answer should have the form $\left(\frac{a}{b}, \frac{c}{d}\right)$.

### Solution:

$x$-coordinate:
\begin{align}
x_m&=\dfrac{x_1+x_2}{2} \\
&=\dfrac{2+5}{2} \\
x_m&=\dfrac{7}{2}
\end{align}

$y$-coordinate:
\begin{align}
y_m&=\dfrac{y_1+y_2}{2} \\
&=\dfrac{-7+18}{2} \\
y_m&=\dfrac{11}{2}
\end{align}

The midpoint is the point $\left(\dfrac{7}{2},\dfrac{11}{2}\right)$.

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### Question 4: Solution Verification
Is $\left(\frac{3}{2}, \frac{4}{3}\right)$ a solution to the equation $8x - 6y = 5$?
Show your work.

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### Question 5: Determine if a Set is a Function
Is the following set of points a function?
$\{(0, 4), (-1, 6), (1, 6), (2, 6), (-2, 3), (2, 5)\}$
Explain your answer.

### Solution:

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### Question 6: Functionality of a Table
Determine if the following table represents a function:

| Domain | Range |
|--------|-------|
| 1 | 3 |
| 2 | 3 |
| 3 | 7 |
| 4 | 3 |
| 5 | 7 |
| 6 | 7 |
| 6 | 3 |
| 8 | 7 |

### Solution:

---

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### Question 7: Graph-Based Functionality
Determine if the following diagram represents a function.
Explain your reasoning.


### Solution:

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### Question 8: Graph Analysis
Using the graph of a function $f(x)$ shown in Figure 1, find the following values:


1. $f(1)=$
2. $f(-2)=$
3. $f(0)=$

### Solution:

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### Question 9: Function Analysis from Graph
Consider the graph in Figure 2. Is it a function?
Explain briefly.


### Solution:

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### Question 10: Evaluate a Rational Function
Given:

$$g(x) = \frac{x+4}{x-2}$$

Find the following values:

1. $g(5)=$
2. $g(-7)=$
3. $g\left(\frac{5}{3}\right)=$

### Solution:

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### Question 11: Graphing a Line
Graph the equation $y = -\frac{1}{2}x - 2$ using the slope and $y$-intercept.


### Solution:

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### Question 12: Equation of a Line
Find the $y=mx+b$ form for the equation of the line through $(3, -2)$ with slope $-\frac{1}{4}$.

### Solution:

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### Question 13: Line Analysis
Consider the line $2x - 7y = 8$.

1. What is its slope?
2. What is the equation of the line perpendicular to this line through $(1, -5)$?

### Solution:

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### Question 14: Parallel Line Analysis
Consider the line $2x - 9y = 8$.

1. What is its slope?
2. What is the equation of the line parallel to this line through $(1, -5)$?

### Solution:

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### Question 15: Slope and Equation of a Line
Consider the points $(2, 4)$ and $(-8, 7)$.

1. What is the slope through these points?
2. What is the equation of the line through these points?

### Solution:

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### Question 16: Parallel Line to $x=-4$
Find the equation of the line through $(4, -2)$ parallel to the line $x = -4$.


### Solution:

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### Question 17: Perpendicular Line to $x=-4$
Find the equation of the line through $(4, -2)$ perpendicular to the line $x = -4$.


### Solution:

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### Question 18: Solve a Linear Equation
Solve:

$$2(2x+1) - 3(-x+5) = 4(3x-1)$$

### Solution:

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### Question 19: Solve Another Linear Equation
Solve:

$$3(2x+1) - 2(x+5) = 4(x-3)$$

### Solution:

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### Question 20: Solve an Inequality
Solve:

$$1 - 2(3x-1) \leq 2x + 5$$

### Solution:

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### Question 21: Domain of a Function
Find the domain of:

$$f(x) = \sqrt{-5x+9}$$

### Solution:

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### Question 22: Increasing, Decreasing, and Extrema
A graph of a function $f(x)$ is shown in Figure 3. Using the graph, determine:


1. Intervals where $f(x)$ is **increasing**.
2. Intervals where $f(x)$ is **decreasing**.
3. Relative **maximum** value.
4. Relative **minimum** value.

### Solution:

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### Question 23: Behavior of a Graph
A graph of a function $f(x)$ is shown in Figure 4. Using the graph, state:


1. Intervals where $f(x)$ is **increasing**.
2. Intervals where $f(x)$ is **decreasing**.
3. Intervals where $f(x)$ is **constant**.

### Solution:

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### Question 24: Sum and Product of Functions
Let $f(x) = x - 5$ and $g(x) = 3x + 7$.
Determine:

1. $(f + g)(x)$
2. $(f g)(x)$

### Solution:

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### Question 25: Composite Functions
Let $f(x) = x^4$ and $g(x) = 2x^5 - 3x^2 + 4$.
Calculate and simplify:

1. $(f g)(x)=$
2. $\left(\dfrac{g}{f}\right)(x)=$

### Solution:

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### Question 26: Composite Functions (Another Example)
Let $f(x) = 2x^5$ and $g(x) = 3x^4 - 2x^3 + 2$.
Calculate and simplify:

1. $\left(\dfrac{f}{g}\right)=$
2. $\left(\dfrac{g}{f}\right)=$

### Solution:

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### Question 27: Nested Composite Functions
Let $f(x) = x^2 + 2$ and $g(x) = 3x - 2$.
Calculate and simplify:

1. $(f \circ g)(x)=$
2. $(g \circ f)(x)=$

### Solution:

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### Question 28: Decompose a Function
Let $h(x) = (7 - 2x)^4$. Find $f(x)$ and $g(x)$ such that $h(x) = (f \circ g)(x)$.
Do not allow $f(x) = x$ or $g(x) = x$.

1. $f(x)=$
2. $g(x)=$

### Solution:

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### Question 29: Decompose a Function (Another Example)
Let:

$$h(x) = \frac{1}{(2x+5)^3}$$

Find $f(x)$ and $g(x)$ such that $h(x) = (f \circ g)(x)$.
Do not allow $f(x) = x$ or $g(x) = x$.

1. $f(x)=$
2. $g(x)=$

### Solution:

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### Question 30: Decompose a Function.
Let $h(x)=\sqrt{\dfrac{2x+1}{3x-5}}$. Find $f(x)$ and $g(x)$ such that $h(x) = (f \circ g)(x)$.
Do not allow $f(x) = x$ or $g(x) = x$.

1. $f(x)=$
2. $g(x)=$

### Solution:

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### Question 31: Solve for the Profit Function
If the revenue from producing $x$ units is $R(x) = 20x - 2x^2$ and the cost from producing $x$ units is $C(x) = 5x + 2$, find the profit function $P(x)$.

### Solution:

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### Question 32: Solve a Quadratic Equation $b=0$
Solve the equation:

$$3x^2 - 15 = 0$$

### Solution:

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### Question 33: Solve Another Quadratic Equation $c=0$
Solve the equation:

$$7x^2 = 4x$$

### Solution:

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### Question 34: Solve by Factoring
Find the zeroes of the function by factoring:

$$f(x) = x^2 - 10x - 24$$

### Solution:

A zero a function is found by setting the entire function itself equal to zero:

\begin{align}
x^2-10x-24&=0 \\
(x-12)(x+2)&=0
\end{align}

Setting each factor equal to zero gives $x-12=0$ and $x+2=0$.
Thus the two solutions are $x=12$ and $x=-2$.

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### Question 35: Solve a Quadratic Equation by Factoring
Solve:

$$2x^2 - 6x = -4$$

### Solution:

Putting it in $ax^2+bx+c=0$ form and dividing both sides by 2 to clean it up:

\begin{align}
2x^2-6x&=-4 \\
2x^2-6x+4&=0 \\
\dfrac{2x^2-6x+4}{2}&=\dfrac{0}{2} \\
x^2-3x+2&=0 \\
(x-2)(x-1)&=0 \\
\end{align}

Setting each factor equal to zero gives $x-2=0$ and $x-1=0$.

Thus the two solutions are $x=2$ and $x=1$.

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### Question 36: Solve Using the Quadratic Formula
Solve the equation using the quadratic formula:

$$2x^2 - 2x - 7 = 0$$

$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

### Solution:

$2x^2 - 2x - 7 = 0$ is in $ax^2+bx+c=0$ form $\checkmark$ , so we can correctly identify $a,b,c$.

$a=2$, $b=-2$, and $c=-7$

#### 1. $b^2-4ac$:

\begin{align}
b^2-4ac&=(-2)^2-4(2)(-7) \\
&=4+(-8)(-7) \\
&=4+56 \\
&=60
\end{align}

#### 2. $\sqrt{b^2-4ac}$ and simplify:

\begin{align}
\sqrt{60}&=\sqrt{4}\sqrt{15}=2\sqrt{15}
\end{align}

#### 3. Finish the rest of the quadratic formula:

\begin{align}
x&=\dfrac{-b \pm 2\sqrt{15}}{2a} \\
x&=\dfrac{-(-2) \pm 2\sqrt{15}}{2(2)} \\
x&=\dfrac{2 \pm 2\sqrt{15}}{4} \\
x&=\dfrac{2}{4} \pm \dfrac{2\sqrt{15}}{4} \\
x&=\dfrac{1}{2} \pm \dfrac{\sqrt{15}}{2}
\end{align}

The two solutions are $x=\dfrac{1}{2}+\dfrac{\sqrt{15}}{2}$ and $x=\dfrac{1}{2}-\dfrac{\sqrt{15}}{2}$

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### Question 37: Solve a Cubic Equation

$$y^3+7y^2+12y=0$$

### Solution:

Factor out the GCF $y$, then factor $y^2+7y+12=(y+3)(y+4)$ (Check by FOIL). Then zero product property.
\begin{align}
y^3+7y^2+12y&=0 \\
y(y^2)+y(7y)+y(12)&=0 \\
y(y^2+7y+12)&=0 \\
y(y+3)(y+4)&=0 \\
\end{align}

Setting each factor equal to zero gives:

$y=0$, $y+3=0$, and $y+4=0$.

Thus $y=0$, $y=-3$, and $y=-4$.

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### Question 38: Solve a Rational Equation
Solve:

$$\dfrac{1}{4}+\dfrac{4}{x}=\dfrac{1}{3}+\dfrac{3}{x}$$

### Solution:

The denominators are $4,x,3$. Thus the LCD is $12x$.

\begin{align}
\dfrac{1}{4}+\dfrac{4}{x}&=\dfrac{1}{3}+\dfrac{3}{x} \\
(12x)\dfrac{1}{4}+(12x)\dfrac{4}{x}&=(12x)\dfrac{1}{3}+(12x)\dfrac{3}{x} \\
3x+48&=4x+36 \\
3x-4x+48&=4x-4x+36 \\
-x+48&=36 \\
-x+48-48&=36-48 \\
-x&=-12 \\
\dfrac{-x}{-1}&=\dfrac{-12}{-1} \\
x&=12
\end{align}

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### Question 39: Solve a Rational Equation

$\dfrac{5}{x-3}=\dfrac{3}{x+3}$

### Solution:

The denominators are $x-3$ and $x+3$. The Lowest common denominator (LCD) is $(x-3)(x+3)$. Multiply both sides by the LCD $(x-3)(x+3)$ and cancel common factors to get rid of all fractions from the equation:

\begin{align}
\dfrac{5}{x-3}&=\dfrac{3}{x+3} \\
(x+3)(x-3)\dfrac{5}{x-3}&=\dfrac{3}{x+3} \cdot (x+3)(x-3) \\
(x+3)(5)&=(3)(x-3) \\
5x+15&=3x-9 \\
5x-3x+15&=3x-3x-9 \\
2x+15&=-9 \\
2x+15-15&=-9-15 \\
2x&=-24 \\
x&=\dfrac{-24}{2} \\
x&=-12
\end{align}

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### Question 40: Solve a Radical Equation. $7+\sqrt{2x-5}=10$

### Solution:

What's happening to x in the expression $7+\sqrt{2x-5}$?

Starting with $x$:
1. Multiplication by 2 to get $2x$
2. Subtraction by 5 to get $2x-5$
3. Square root to get $\sqrt{2x-5}$
4. Add 7 to get $+7+\sqrt{2x-5}$

**To undo this, we do the opposite things in reverse order:**

1. Subtract by 7
2. Square
3. Add by 5
4. Division by 2.

\begin{align}
7+\sqrt{2x-5}&=10 \\
7-7+\sqrt{2x-5}&=10-7 \qquad \text{Subtract 7} \\
\sqrt{2x-5}&=3 \\
(\sqrt{2x-5})^2&=(3)^2 \qquad \text{Square}\\
2x-5&=9 \\
2x-5+5&=9+5 \qquad \text{Add 5} \\
2x&=14 \\
\dfrac{2x}{2}&=\dfrac{14}{2} \qquad \text{Divide by 2} \\
x&=7
\end{align}

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### Question 41: Analyze End Behavior of a Polynomial
Analyze the end behavior of:

1. $f(x) = -3x^4 - 2x^2 - x + 4$

2. $g(x) = 20x - 2x^3 + 4x^5$


### Solution:

#### 1.

The leading term of $f(x)$ is $-3x^4$. The leading coefficient is $-3$. The degree is $4$.

Negative leading coefficient and even degree means it falls to the left and falls to the right. The 3rd choice is correct.

#### 2.

The leading term of $g(x)$ is $4x^5$. The leading coefficient is $4$. The degree is $5$.

Positive leading coefficient and odd degree means it falls to the left and rises to the right. The 2nd choice is correct.

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### Question 42: Solve a Polynomial Inequality
Solve:

$$(x-2)(x+4) \geq 0$$

### Solution:

#### 1. Find partition numbers by setting the polynomial equal to zero.

$(x-2)(x+4)=0$ means $x-2=0$ or $x+4=0$. Thus $x=2$ and $x=-4$.

Thus $(x-2)(x+4)$ is zero when $x=2$ and $x=-4$. Include both endpoints.

#### 2. These two partition numbers $2$ and $-4$ partitions the number line into three subintervals $(-\infty,-4)$ , $(-4,2)$, and $(2,\infty)$.

Pick a test number in each subinterval and test it.

| Subinterval | Test value $x$ | Plug in $(x-2)(x+4)$ | Answer | Sign |
|---|---|---|---|---|
|$(-\infty,-4)$ | $x=-5$ | $((-5)-2)((-5)+4)$ | $7$ | **Positive** |
|$(-4,2)$ | $x=0$ | $((0)-2)((0)+4)$ | $-\dfrac{1}{2}$ | **Negative** |
|$(2,\infty)$ | $x=3$ | $((3)-2)((3)+4)$ | $\dfrac{1}{7}$ | **Positive** |

#### 3. Conclusion.

$(x-2)(x+4) \leq 0$ means we want to find when $(x-2)(x+4)$ is **negative** and **zero**.

By the above table, $(x-2)(x+4)$ is negative in the interval $\left(-4,2\right)$.

$(x-2)(x+4)$ is equal to zero when $x=-4$ and $x=2$ so include both endpoints.

Thus our final answer in interval notation is $\left[-4,2\right]$

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### Question 43. Let $k(x)=-3x^4(x+2)$.

1. Find the zeros and state the multiplicity of each.
2. What is the leading term of $k(x)$?
3. Draw a rough graph, illustrating end behavior, and behavior at intercepts (cross or bounce).

### Solution:

#### 1. The following table gives the factors, zeros, and multiplicities.

| Factor | Zero | Multiplicity |
|---|---|---|
|$-3$ | N/A | N/A |
|$x^4$ | $x=0$ | $4$ |
| $(x+2)$ | $x=-2$ | $1$ |

$x=0$ has multiplicity 4.
$x=-2$ has multiplicity 1.

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### Question 44: Vertical and Horizontal Asymptotes
For the function:

$$g(x) = \frac{3x^2 + 2}{x^2 + 5x}$$

1. Find the vertical asymptotes.
2. Find the horizontal asymptotes.

### Solution:

#### 1. Vertical asymptotes are found by setting the denominator equal to zero:

\begin{align}
x^2+5x&=0 \\
x(x)+x(5)&=0 \\
x(x+5)&=0 \\
x=0 \quad &\text{or} \quad x+5=0 \\
x=0 \quad &\text{or} \quad x=-5 \\
\end{align}

The function $g(x)$ has two vertical asymptotes $x=0$ and $x=-5$.

#### 2. Horizontal asymptotes are found by comparing the degrees of top and bottom.

The top is $3x^2+2$ which is degree 2.
The bottom is $x^2+5x$ which is also degree 2.

The top and bottom degrees are equal. Then throw away lesser order terms.

$3x^2$ turns into $3x^2$.

$x^2+5x$ turns into $x^2$.

$\dfrac{3x^2+2}{x^2+5x} \to \dfrac{3x^2}{x^2}=3$

Thus the function has a horizontal asymptote $y=3$.

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### Question 45: Rational Inequality. Solve $\dfrac{x+6}{3x-4} \leq 0$

### Solution:

#### 1. Find partition numbers by setting top and bottom equal to zero.

| Top=0 | Bottom=0 |
|---|---|
|$x+6=0$ | $3x-4=0$ |
|$x=-6$ | $3x=4$ |
|$x=-6$ | $x=\dfrac{4}{3}$ |

Thus $\dfrac{x+6}{3x-4}$ is zero when $x=-6$.

$\dfrac{x+6}{3x-4}$ is undefined when $x=4/3$. Exclude $4/3$.

#### 2. These two partition numbers $-6$ and $4/3$ partitions the number line into three subintervals $(-\infty,-6)$ , $(-6,4/3)$, and $(4/3,\infty)$.

Pick a test number in each subinterval and test it.

| Subinterval | Test value $x$ | Plug in $\dfrac{x+6}{3x-4}$ | Answer | Sign |
|---|---|---|---|---|
|$(-\infty,-6)$ | $x=-7$ | $\dfrac{-7+6}{3(-7)-4}$ | $\dfrac{1}{25}$ | **Positive** |
|$(-6,4/3)$ | $x=0$ | $\dfrac{0+6}{3(0)-4}$ | $-\dfrac{6}{4}$ | **Negative** |
|$(4/3,\infty)$ | $x=2$ | $\dfrac{2+6}{3(2)-4}$| $4$ | **Positive** |

#### 3. Conclusion.

$\dfrac{x+6}{3x-4} \leq 0$ means we want to find when $\dfrac{x+6}{3x-4}$ is **negative** and **zero**.

By the above table, $\dfrac{x+6}{3x-4}$ is negative in the interval $\left(-6,\dfrac{4}{3}\right)$.

$\dfrac{x+6}{3x-4}$ is equal to zero when the numerator is $-6$, we include $-6$.

Thus our final answer in interval notation is $\left[-6,\dfrac{4}{3}\right)$

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### Question 46: Graphing a Rational Function
For the function:

$$f(x) = \frac{2x+2}{3x-6}$$

1. Find the $x$-intercepts.
2. Find the $y$-intercept.
3. Find the vertical asymptotes.
4. Find the horizontal asymptotes.
5. Graph $f(x)$.

### Solution:

#### 1. $x$-intercept is when $y=0$, so set the entire function itself equal to zero and solve:

\begin{align}
\dfrac{2x+2}{3x-6}&=0 \\
(3x-6)\cdot \dfrac{2x+2}{3x-6}&=0\cdot (3x-6) \\
2x+2&=0 \\
2x&=-2 \\
x&=-1
\end{align}

The $x$-intercept is the point $(-1,0)$.

#### 2. The $y$-intercept is when $x=0$:

$$f(x) = \frac{2x+2}{3x-6}$$
$$f(0) = \frac{2(0)+2}{3(0)-6}=\dfrac{2}{-6}=-\dfrac{1}{3}$$

The $y$-intercept is the point $\left(0,-\dfrac{1}{3}\right)$

#### 3. Vertical Asymptotes are when the denominator is equal to zero:

$f(x) = \frac{2x+2}{3x-6} \rightarrow 3x-6=0$

$3x=6$

$x=2$

$f(x)$ has a vertical asymptote $x=2$.

#### 4. Horizontal asymptotes are found by comparing the degrees of the top and bottom:

$f(x) = \frac{2x+2}{3x-6}$

The top $2x+2$ is degree 1.
The bottom $3x-6$ is degree 1.
Since the degrees are equal, the horizontal asymptote is found by throwing away all the lesser order terms.

$2x+2$ turns into $2x$.

and

$3x-6$ turns into $3x$

$f(x) = \frac{2x+2}{3x-6} \to \dfrac{2x}{3x}=\dfrac{2}{3}$.

Thus the function has a horizontal asymptote $y=\dfrac{2}{3}$.

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### Question 47. Graph the function by substituting and plotting points: $f(x)=-2^x$

### Solution:

| $x$ | $y=f(x)=-2^x$ | $(x,y)$ |
|---|---|---|
|$-3$ | $-2^{-3}=-\dfrac{1}{2^3}=-\dfrac{1}{8}$ | $\left(-3,-\dfrac{1}{8}\right)$ |
|$-2$ | $-2^{-2}=-\dfrac{1}{2^2}=-\dfrac{1}{4}$ | $\left(-3,-\dfrac{1}{4}\right)$ |
|$-1$ | $-2^{-1}=-\dfrac{1}{2^1}=-\dfrac{1}{2}$ |$\left(-3,-\dfrac{1}{2}\right)$ |
|$0$ | $-2^{0}=-1$ | $(0,-1)$ |
|$1$ | $-2^{1}=-2$ | $(1,-2)$ |
|$2$ | $-2^{2}=-4$ | $(2,-4)$ |
|$3$ | $-2^{3}=-8$ | $(3,-8)$ |

Plotting all those points on a grid and connecting the dots:

[graph goes here]

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### Question 48: Logarithmic Evaluations
Evaluate:

1. $\log_2(16)=$
2. $\log_4(1)=$
3. $\log_2\left(\frac{1}{4}\right)=$
4. $\log_5 5^4=$

### Solution:

1. $\log_2(16)=4$ since $2^4=16$
2. $\log_4(1)=0$ since $4^0=1$
3. $\log_2\left(\frac{1}{4}\right)=-2$ since $2^{-2}=\dfrac{1}{4}$
4. $\log_5 5^4=4$ since $5^4=5^4$

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### Question 49: Converting between Exponential and Logarithmic Equations

1. $4^t=5$
2. $\log_4 (y)=3$

### Solution:

1. $\log_4 (5)=t$
2. $4^3=y$
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### Question 50: Compound Interest
If the principal is $$1000$, compounded quarterly at an annual rate of $8\%$ for $4$ years, what is the final amount?

1. $P=$
2. $r=$
3. $n=$
4. $t=$
5. $A=$

### Solution:
1. $P=1000$
2. $r=0.08$
3. $n=4$
4. $t=4$
5.

\begin{align}
A&=P\left(1+\dfrac{r}{n}\right)^{nt} \\
&=1000\left(1+\dfrac{0.08}{4}\right)^{4(4)} \\
&=1000(1+0.02)^{16} \\
&=1000(1.02)^{16} \\
&\approx 1372.79
\end{align}

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### Question 52: Change to Logarithmic Form
Change to logarithmic form and solve for $x$:

$$4^{6x} = 36$$

### Solution:

\begin{align}
4^{6x} &= 36 \\
\log_4 (36)&=6x \\
6x&=\log_4 (36) \\
x&=\dfrac{\log_4(36)}{6}
\end{align}

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### Question 53: Change to Exponential Form
Change to exponential form and solve for $x$:

$$\log_2(3x+1) = 2$$

#### Solution:

\begin{align}
\log_2(3x+1)&=2 \\
2^2&=3x+1 \\
4&=3x+1 \\
3&=3x \\
3x&=3 \\
x&=1
\end{align}

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