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:

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

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

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

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

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

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

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

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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:

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

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Question 2: Numerical Evaluation

Evaluate the following:

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

Solution:

1. $(193.76)(100)=19376$
2. $0.06763÷100=0.0006763$
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

$(\frac{a}{b},\frac{c}{d})$.

Solution:

$x$-coordinate:

$$\begin{array}{rl}{x}_m& ={\displaystyle \frac{x_1+x_2}{2}}\\ & ={\displaystyle \frac{2+5}{2}}\\ x_m& ={\displaystyle \frac{7}{2}}\end{array}$$

$y$-coordinate:

$$\begin{array}{rl}{y}_m& ={\displaystyle \frac{y_1+y_2}{2}}\\ & ={\displaystyle \frac{-7+18}{2}}\\ y_m& ={\displaystyle \frac{11}{2}}\end{array}$$

The midpoint is the point

$({\displaystyle \frac{7}{2}},{\displaystyle \frac{11}{2}})$.

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Question 4: Solution Verification

Is

$(\frac{3}{2},\frac{4}{3})$ 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:

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

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. $(fg)(x)$

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. $(fg)(x)=$
2. $\left({\displaystyle \frac{g}{f}}\right)(x)=$

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. $\left({\displaystyle \frac{f}{g}}\right)=$
2. $\left({\displaystyle \frac{g}{f}}\right)=$

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

$h(x)=\sqrt{{\displaystyle \frac{2x+1}{3x-5}}}$. 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:

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

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

$a{x}^2+bx+c=0$ form and dividing both sides by 2 to clean it up:

$$\begin{array}{rl}2x^2-6x& =-4\\ 2x^2-6x+4& =0\\ {\displaystyle \frac{2x^2-6x+4}{2}}& ={\displaystyle \frac{0}{2}}\\ x^2-3x+2& =0\\ (x-2)(x-1)& =0\end{array}$$

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:

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

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

Solution:

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

$a{x}^2+bx+c=0$ form

$✓$ , so we can correctly identify

$a,b,c$.

$a=2$,

$b=-2$, and

$c=-7$

1.

$b^2-4ac$:

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

2.

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

$$\begin{array}{rl}\sqrt{60}& =\sqrt{4}\sqrt{15}=2\sqrt{15}\end{array}$$

3. Finish the rest of the quadratic formula:

$$\begin{array}{rl}x& ={\displaystyle \frac{-b\pm 2\sqrt{15}}{2a}}\\ x& ={\displaystyle \frac{-(-2)\pm 2\sqrt{15}}{2(2)}}\\ x& ={\displaystyle \frac{2\pm 2\sqrt{15}}{4}}\\ x& ={\displaystyle \frac{2}{4}}\pm {\displaystyle \frac{2\sqrt{15}}{4}}\\ x& ={\displaystyle \frac{1}{2}}\pm {\displaystyle \frac{\sqrt{15}}{2}}\end{array}$$

The two solutions are

$x={\displaystyle \frac{1}{2}}+{\displaystyle \frac{\sqrt{15}}{2}}$ and

$x={\displaystyle \frac{1}{2}}-{\displaystyle \frac{\sqrt{15}}{2}}$

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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.

$$\begin{array}{rl}{y}^3+7y^2+12y& =0\\ y(y^2)+y(7y)+y(12)& =0\\ y(y^2+7y+12)& =0\\ y(y+3)(y+4)& =0\end{array}$$

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:

$${\displaystyle \frac{1}{4}}+{\displaystyle \frac{4}{x}}={\displaystyle \frac{1}{3}}+{\displaystyle \frac{3}{x}}$$

Solution:

The denominators are

$4,x,3$. Thus the LCD is

$12x$.

$$\begin{array}{rl}{\displaystyle \frac{1}{4}}+{\displaystyle \frac{4}{x}}& ={\displaystyle \frac{1}{3}}+{\displaystyle \frac{3}{x}}\\ (12x){\displaystyle \frac{1}{4}}+(12x){\displaystyle \frac{4}{x}}& =(12x){\displaystyle \frac{1}{3}}+(12x){\displaystyle \frac{3}{x}}\\ 3x+48& =4x+36\\ 3x-4x+48& =4x-4x+36\\ -x+48& =36\\ -x+48-48& =36-48\\ -x& =-12\\ {\displaystyle \frac{-x}{-1}}& ={\displaystyle \frac{-12}{-1}}\\ x& =12\end{array}$$

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Question 39: Solve a Rational Equation

${\displaystyle \frac{5}{x-3}}={\displaystyle \frac{3}{x+3}}$

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:

$$\begin{array}{rl}{\displaystyle \frac{5}{x-3}}& ={\displaystyle \frac{3}{x+3}}\\ (x+3)(x-3){\displaystyle \frac{5}{x-3}}& ={\displaystyle \frac{3}{x+3}}\cdot (x+3)(x-3)\\ (x+3)(5)& =(3)(x-3)\\ 5x+15& =3x-9\\ 5x-3x+15& =3x-3x-9\\ 2x+15& =-9\\ 2x+15-15& =-9-15\\ 2x& =-24\\ x& ={\displaystyle \frac{-24}{2}}\\ x& =-12\end{array}$$

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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.

$$\begin{array}{rl}7+\sqrt{2x-5}& =10\\ 7-7+\sqrt{2x-5}& =10-7\text{Subtract 7}\\ \sqrt{2x-5}& =3\\ (\sqrt{2x-5}{)}^2& =(3{)}^2\text{Square}\\ 2x-5& =9\\ 2x-5+5& =9+5\text{Add 5}\\ 2x& =14\\ {\displaystyle \frac{2x}{2}}& ={\displaystyle \frac{14}{2}}\text{Divide by 2}\\ x& =7\end{array}$$

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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:

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

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

$(-\mathrm{\infty },-4)$ ,

$(-4,2)$, and

$(2,\mathrm{\infty })$.

Pick a test number in each subinterval and test it.

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

3. Conclusion.

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

$(-4,2)$.

$(x-2)(x+4)$ is equal to zero when

$x=-4$ and

$x=2$ so include both endpoints.

Thus our final answer in interval notation is

$[-4,2]$

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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:

$$\begin{array}{rl}{x}^2+5x& =0\\ x(x)+x(5)& =0\\ x(x+5)& =0\\ x=0& \text{or}x+5=0\\ x=0& \text{or}x=-5\end{array}$$

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

$x^2$.

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

Thus the function has a horizontal asymptote

$y=3$.

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Question 45: Rational Inequality. Solve

${\displaystyle \frac{x+6}{3x-4}}\le 0$

Solution:

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

Top=0	Bottom=0
$x+6=0$	$3x-4=0$
$x=-6$	$3x=4$
$x=-6$	$x={\displaystyle \frac{4}{3}}$

Thus

${\displaystyle \frac{x+6}{3x-4}}$ is zero when

$x=-6$.

${\displaystyle \frac{x+6}{3x-4}}$ 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

$(-\mathrm{\infty },-6)$ ,

$(-6,4/3)$, and

$(4/3,\mathrm{\infty })$.

Pick a test number in each subinterval and test it.

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

3. Conclusion.

${\displaystyle \frac{x+6}{3x-4}}\le 0$ means we want to find when

${\displaystyle \frac{x+6}{3x-4}}$ is negative and zero.

By the above table,

${\displaystyle \frac{x+6}{3x-4}}$ is negative in the interval

$(-6,{\displaystyle \frac{4}{3}})$.

${\displaystyle \frac{x+6}{3x-4}}$ is equal to zero when the numerator is

$-6$, we include

$-6$.

Thus our final answer in interval notation is

$[-6,{\displaystyle \frac{4}{3}})$

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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:

$$\begin{array}{rl}{\displaystyle \frac{2x+2}{3x-6}}& =0\\ (3x-6)\cdot {\displaystyle \frac{2x+2}{3x-6}}& =0\cdot (3x-6)\\ 2x+2& =0\\ 2x& =-2\\ x& =-1\end{array}$$

The

$x$-intercept is the point

$(-1,0)$.

2. The

$y$-intercept is when

$x=0$:

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

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

The

$y$-intercept is the point

$(0,-{\displaystyle \frac{1}{3}})$

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

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

$3x=6$

$x=2$

$f(x)$ 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

$3x$

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

Thus the function has a horizontal asymptote

$y={\displaystyle \frac{2}{3}}$.

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Question 47. Graph the function by substituting and plotting points:

$f(x)=-2^x$

Solution:

$x$	$y=f(x)=-2^x$	$(x,y)$
$-3$	$-2^{-3}=-{\displaystyle \frac{1}{2^3}}=-{\displaystyle \frac{1}{8}}$	$(-3,-{\displaystyle \frac{1}{8}})$
$-2$	$-2^{-2}=-{\displaystyle \frac{1}{2^2}}=-{\displaystyle \frac{1}{4}}$	$(-3,-{\displaystyle \frac{1}{4}})$
$-1$	$-2^{-1}=-{\displaystyle \frac{1}{2^1}}=-{\displaystyle \frac{1}{2}}$	$(-3,-{\displaystyle \frac{1}{2}})$
$0$	$-2^0=-1$	$(0,-1)$
$1$	$-2^1=-2$	$(1,-2)$
$2$	$-2^2=-4$	$(2,-4)$
$3$	$-2^3=-8$	$(3,-8)$

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
   $2^{-2}={\displaystyle \frac{1}{4}}$
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

$8\mathrm{\%}$ 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. $t=4$

$$\begin{array}{rl}A& =P{(1+{\displaystyle \frac{r}{n}})}^{nt}\\ & =1000{(1+{\displaystyle \frac{0.08}{4}})}^{4(4)}\\ & =1000(1+0.02{)}^{16}\\ & =1000(1.02{)}^{16}\\ & \approx 1372.79\end{array}$$

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Question 52: Change to Logarithmic Form

Change to logarithmic form and solve for

$x$:

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

Solution:

$$\begin{array}{rl}{4}^{6x}& =36\\ {\log }_4(36)& =6x\\ 6x& ={\log }_4(36)\\ x& ={\displaystyle \frac{{\log }_4(36)}{6}}\end{array}$$

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Question 53: Change to Exponential Form

Change to exponential form and solve for

$x$:

$${\log }_2(3x+1)=2$$

Solution:

$$\begin{array}{rl}{\log }_2(3x+1)& =2\\ 2^2& =3x+1\\ 4& =3x+1\\ 3& =3x\\ 3x& =3\\ x& =1\end{array}$$

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