## Comparing Data in Different Formats

This post about comparing data in different formats is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

Liam’s Pay

This graph shows how much Liam gets paid as a function of the number of hours he works. Sarah, on the other hand, is paid according to the formula S=(75/4)h, where S represents Sarah’s pay and h represents the number of hours she has worked.

Is Liam’s rate of pay more or less than Sarah’s, and by how much?

## Solution

You’re being asked to compare data that measure similar things but are given to you in different formats.

The trick is to convert one of those expressions into the same format as the other. Or, if you can’t do that, then convert both to some third format.

P=(75/4)h looks . . . almost straightforward. That fraction may be inconvenient. 75/4 = 18.75, so let’s say S = 18.75h represents Sarah’s rate of pay.

Now look at the graph. Can you represent Liam’s pay as something like L=kh, where k is Liam’s rate of pay? Since the graph goes through the origin, its equation will be of the form L=something*and “something” is the slope. To find it from the graph, pick two points on the line that are easy to read, and calculate. The origin — with coordinates (0, 0) — looks easy. And the point (2, 40) doesn’t look bad. The slope is the change in y divided by the change in for those two points:

$\fn_jvn&space;\frac{40-0}{2-0}=\frac{40}{2}=20$

So Liam’s pay is L=20h.

You’re asked for the difference between Liam’s rate of pay and Sarah’s. That is

$\fn_jvn&space;\20.00/hr&space;-&space;\18.75/hr&space;=&space;\1.25/hr$

Answer: Liam gets paid \$1.25 per hour more than Sarah.

For more about relating a line’s equation to its graph, see “Finding a Line’s Equation from the Graph of Its Perpendicular.

This question is similar to question number 5 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

## Finding a Line’s Equation from the Graph of Its Perpendicular

This is the line whose perpendicular you need to find.

This post about finding the equation of a line that is perpendicular to a line you read from a graph is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

Find the equation of the line that goes through the origin and is perpendicular to the blue line shown.

## Solution

The origin is the point on the coordinate plane with the coordinates $\inline \fn_jvn (0, 0)$.

You are asked to find the equation of the line that is perpendicular to the one shown and passes through the origin.

### Slope

How do you find the equation of a perpendicular line? Start with the slope: If two lines are perpendicular and their slopes are  $\fn_jvn m_1$  and  $\fn_jvn m_2$, then $\fn_jvn m_2=-1/m_1$. One is the negative reciprocal of the other. Another way to say this: $\fn_jvn m_1\cdot m_2=-1$.

Next, find  $\fn_jvn m_1$, the slope of the line shown. Then the slope of the line whose equation we seek equals $\fn_jvn -1/m_1$.

To do that, choose two points with coordinates you can read. The intercepts — (-2, 0) and (0, -2) — suggest themselves. The slope of the line between those points is the difference between their y-values divided by the difference between their x-values. The difference in y-values is $\fn_jvn -2-0=-2$. The difference in x-values is $\fn_jvn 0-(-2)=2$. Then the slope of the line between the points is:

$\fn_jvn&space;{\color{DarkBlue}&space;\frac{-2}{2}=-1}$

So the original line has a slope $\inline \fn_jvn m_1=-1$. Then the line we are looking for has a slope  $\inline \fn_jvn m_2=-1/m_1=-1/-1=1$.

### Intercept

From there, all we need for the line’s equation is the y-intercept. You’re told the line passes through the point (0, 0), so the y-intercept — the b in $\inline&space;\fn_phv&space;y=mx+b$  — is 0.

A line that passes through the origin has a y-intercept of 0.

Then this line’s y-intercept is 0 and its equation is

$\fn_jvn {\color{DarkBlue} y=x}$

That’s the brown line on the graph below.

The brown line is a perpendicular that passes through the origin.

The brown line looks perpendicular to the blue one and goes through the origin. Looks about right.

For more about relating a line’s equation to its graph, see “Comparing Data in Different Formats.

This question is similar to question number 2 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

## How to Solve an Exponential Equation

This post about how to solve an exponential equation is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

Solve for x:

$\fn_jvn 5^{3x}=6$

## Solution

You are asked to solve this exponential equation.

How do you get that x out of the exponent so you can solve for it? Consider the relationship between exponents and logarithms:

$\fn_jvn a^b=c<=>\log_a c=b$

where “<=>” means the equations on both sides of the symbol are equivalent; each is true only when the other is true.

Note that the equation on the right-hand side is solved for b, the exponent. Let’s apply that to the equation we’re given. 5 stands in the place of a, 3x in the place of b, and 6 in the place of c. Substituting those values into $\inline \fn_jvn \log_a c=b$ gives:

$\fn_jvn \log_5 6=3x$

Divide both sides by 3 and switch the sides:

$\fn_jvn x=\frac{\log_5 6}{3}$

### Caution

Don’t try to cancel the 6 in the numerator with the 3 in the denominator. $\fn_phv x=\log_{5}6$ is a function carried out on the number 6; it is not a multiple of 6, so it does not cancel with 3.

## Another Way to Solve an Exponential Equation with Logs

Take the log of each side.

 $\inline \fn_jvn 5^{3x}=6$$\inline \fn_jvn 5^{3x}=6$ This is the equation that you have to solve for x. $\inline \fn_jvn \log_{5}5^{3x}=\log_{5}6$$\inline \fn_jvn \log_{5}5^{3x}=\log_{5}6$ If two things are equal, their logs are equal. Why log base 5? Any base is valid; choosing the base that’s in the problem will likely give you the cleanest answer. $\inline \fn_jvn 3x\log _5 5= \log_5 6$$\inline \fn_jvn 3x\log _5 5= \log_5 6$ The power rule for logarithms says $\inline \fn_jvn \log a^b=b\log a$$\inline \fn_jvn \log a^b=b\log a$. In other words, you can bring that exponent down. Apply this rule to the equation’s left-hand side: $\inline \fn_jvn \log _5 5^{3x}=3x\log _5 5$$\inline \fn_jvn \log _5 5^{3x}=3x\log _5 5$ $\inline \fn_jvn 3x \cdot 1=\log_5 6$$\inline \fn_jvn 3x \cdot 1=\log_5 6$ $\inline \fn_jvn \log _a a= 1$$\inline \fn_jvn \log _a a= 1$ so $\inline \fn_jvn \log _5 5= 1$$\inline \fn_jvn \log _5 5= 1$. Substitute 1 for $\inline \fn_jvn \log _5 5$$\inline \fn_jvn \log _5 5$. $\inline \fn_jvn x=\log_5 6/3$$\inline \fn_jvn x=\log_5 6/3$ Divide both sides by 3 and there’s the answer.

Related: “Exponential Growth.”

This question is similar to question number 18 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

## Exponential Growth

This post about exponential growth is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

Last month, a rumor started on the Internet. Today 1000 people believe it. The number of rumor believers doubles every 2 days. Write an equation for the number of rumor believers, n, x days after today.

## Solution

The number of believers increases in proportion to the number of believers already in existence. That is exponential growth.

How can we set up a formula?

Every multiplication by 2 means taking 2 to one more power. A multiplication by 2 happens every 2 days, so 2 is taken to the power of half the number of days that have passed : x/2. That means 2 needs to the power of x/2. Then multiply  by the baseline number, 1000:

$\fn_jvn {\color{Blue} n=1000 \cdot 2^{x/2}}$

To check, try a couple of points where you can calculate both with and without the formula. Choose small numbers to make it easy. Choose even numbers to avoid fractions in formulas. Let’s see what happens 2 days from today, when $\inline \fn_jvn x=2$, and 4 days from today, when $\inline \fn_jvn x=4$.

After 2 days, the number of believers should have doubled from today’s number. So at $\inline \fn_jvn x=2$, n=2000.

Per the formula, at $\inline \fn_jvn x=2$, $\inline \fn_jvn n=1000 \cdot 2^{2/2}=1000 \cdot 2 = 2000$.

After 4 days, the number of believers should have doubled from the number 2 days from today, which itself is double today’s number. So after 4 days, the number of believers should have quadrupled and at $\inline \fn_jvn x=4$, $\inline \fn_jvn n=4000$.

Per the formula, at $\inline \fn_jvn x=4$, $\inline \fn_jvn n=1000 \cdot 2^{4/2}=1000 \cdot 2^2=1000 \cdot 4 = 4000$.

So the equation $\inline \fn_jvn n=1000\cdot 2^{x/2}$ checks out.

Generally, the format of exponential growth equations looks something like  $\inline \fn_jvn N(t)=N_0b^{kt}$, with the letters changing depending on the application. Here, N(t) represents the number of something present at time t, N0 represents the amount present at time t=0, b is the exponential base (2 in the rumor example, where it stands for doubling), and k is a constant that depends on the rate of growth. Another example: For an investment with interest compounded continuously, the formula is $\inline \fn_jvn A(t)=A_0e^{rt}$, where A is the amount present at time t, A0 is the amount present at time t=0, and r is the rate of interest, expressed as a decimal.

As you can see, you can work some problems of exponential growth, like this one, without using the formula.

## Further Thought

Note that the problem tells you there are 1000 rumor-believers today. Since the number doubles every 2 days, there must have been 500 believers the day before yesterday, and 250 two days before that. When did all this start? One way to find out is to keep dividing by 2 until you get to 1, representing that one person who started it all.

 Day Number of Rumor Believers today 1000 2 days ago 1000/2=500 4 days ago 500/2=250 6 days ago 250/2=125 8 days ago 125/2, about 63 10 days ago 63/2, about 32 12 days ago 32/2 = 16 14 days ago 16/2=8 16 days ago 8/2=4 18 days ago 4/2=2 20 days ago 2/2=1

There was some rounding, but it looks like it took about 20 days to get from one person to 1000.

To confirm that result, start with 1 rumor believer and keep doubling that number until you get to 1000. How many doublings does it take? 10 doublings get you to 1024, close enough. And with a doubling every 2 days, 10 doublings take 20 days. For a related problem, see Exponential Decay.

Interesting factoid: To work with an equation for exponential growth or decay, you don’t have to know the starting point; you just need to know data at some known point.

Related: How to Solve an Exponential Equation.

The Accuplacer sample problem on which this one is based is multiple choice. It is #8 in the sample questions for the Accuplacer Advanced Algebra and Functions test.