Newton’s Clock, Part 2

Well I didn’t get very far around Newton’s clock last time because the expression for “1” took a while to explain. Today I will explain the expressions for “2” and “3” on the clock:

The expression log10(100) is a logarithm (logs). I’ve talked about logs before. They are exponents. This particular one is the english equivalent of asking “What is the exponent of 10 so that 10x = 100?”. Well, the answer to that is “2” because 10Β² = 100.

The next expression

\[
\mathop{\int}\limits_{1}\limits^{2}{2xdx}
\]

again, will take a bit of explaining.

This is a calculus expression. The ∫ symbol is an elongated “S”. It has a German origin but this symbol was used because the expression represents a sum (addition) of infinitesimal (that is, ridiculously small) things. For this particular expression, you can think of it as finding the area under the plot of the equation y = 2x from x = 1 to x = 2:

So this expression can be thought of as adding the areas of infinitesimally thin rectangles from x = 1 to x = 2, with a width of dx and a height of 2x. This sum will equal the total area under y = 2x from x = 1 to x = 2. Using calculus, this area is equal to 2Β² – 1Β² = 4 – 1= 3. So this is why this expression is in the “3” position.

I will leave this as an exercise for the reader to confirm that this is the correct area by using geometric formulas for area for triangles or trapeziums (trapezoids in the USA).

Newton’s Clock, Part 1

As I mentioned many posts before, maths is a language – a much more elegant and concise language when taking about maths stuff. This will be very apparent when I talk about the expressions in the clock I just bought for my office:

Each one of the maths expressions on the clock equal the number corresponding to its position on the clock. So I will be talking about each of these.

By far, the expression corresponding to “1” on the clock, illustrates the conciseness of maths expressions:

\[
{e}^{\mathit{\pi}{i}}\cos\mathit{\pi}\hspace{0.33em}{=}\hspace{0.33em}{1}
\]

This expression makes use of more mathematical knowledge than any of the other expressions. It uses the maths constants e and πœ‹, and uses radians, trigonometric (or circular) functions, exponentials, and Euler’s identity. The two parts of this expression are eπœ‹i and cos πœ‹. Let me first talk about cos πœ‹.

Now I have talked about trig functions before, but I mainly talked about sine of an angle (abbreviated “sin”). The cosine of an angle (abbreviated “cos”) is similar. You can take the cosine of an angle on your calculator, but you need to tell the calculator whether you are measuring angles in “degrees” or “radians”. In this expression, we are taking the cosine of πœ‹ radians. If your calculator is in radians mode and if you have ‘πœ‹’ button, taking the cosine of πœ‹ will show -1. If you do not have a πœ‹ button, take the cosine of 3.14179 and you will get an answer approximately equal to -1. So, cos πœ‹ = -1.

Now the eπœ‹i part requires a bit more of an explanation. The numbers e and πœ‹ have been discussed before. They are both irrational numbers which means that you cannot write them down exactly using numbers. We just agree that the symbol e is exactly e and the symbol πœ‹ is exactly πœ‹. If you have an ex button on your calculator, you can get an approximate value for e by typing 1, then the ex key. I have talked about e before. The letter e is for Euler who did a lot of work with this constant. So what is i?

The number i is not a real number. It is actually called an imaginary number. It is formally defined in terms of its square:

i2 = -1

which means that i is the square root of -1. I’m not making this up.

Since in the realm of real numbers, you know that you cannot take the square root of a negative number. So defining i is creating a whole new realm of numbers called complex numbers. Actually, complex numbers are the union of two realms: the reals and the imaginaries. There is a lot more to say about this but this will have to be a topic for a future post.

So what is eπœ‹i where i is in the exponent? There is an identity (a maths rule) called Euler’s identity that explains what eπœ‹i is equal to:

eπœ‹i + 1 = 0

So eπœ‹i must equal -1 for Euler’s identity to work. Again, explaining anything more is the topic of several future posts.

So if eπœ‹i = -1 and cos πœ‹ = -1, then

eπœ‹icos πœ‹ = (-1)(-1) = 1 which is the 1 o’clock spot on the clock.

I haven’t even scratched the surface of explaining this expression, but it has already filled up this post. The other expressions will not take as long to explain.

Logarithms, Part 5

I never thought these posts would get to 5.

Now I said I would do a population problem but I have decided to go with a radioactive decay problem instead. I will use the example of carbon dating as this is based on radioactive decay. But first, let’s look at the general equation for exponential decay:

\[
{A}\hspace{0.33em}{=}\hspace{0.33em}{A}_{0}{e}^{{-}{kt}}
\]

This formula gives the amount of something that is decreasing exponentially. A is the amount left after t seconds, hours, days, years or whatever depending on the value of the rate of decrease factor which is k. A0 is the amount of something we started out with, the amount present at t = 0. This formula makes sense when you look at the ekt part of the equation.

Now I have talked about e before. It is an irrational number, like πœ‹, and is approximately equal to 2.7183. k is a rate of decrease factor that depends on the material we are working with and the units of time t. The –kt part, the exponent of e, is a negative number since k and t are positive. I have explained negative exponents before but ekt equals 1/ekt . Now what happens to ekt as t gets large? Any number greater than 1 (which e is) raised to a larger and larger power, gets very big. And when you divide a big number into 1, you get a very small number. So A0 is being multiplied by a number that gets smaller and smaller as time goes on. That is why A, the amount of material, is exponentially decreasing.

With that as a background, let’s talk about carbon dating. Any living thing has carbon in it. Indeed, all life on earth is carbon-based which means that the the molecules essential for life are composed of lots of carbon. Now carbon comes in different “flavors”. These flavors are called isotopes and carbon has two main isotopes: carbon 12 the most abundant and non-radioactive, and carbon 14 which is radioactive. Fortunately, the amount of carbon 14 is very small – about 1 atom to every 1012 atoms of carbon 12. However, in living things, this ratio is pretty much constant since carbon 14 is continually made in our atmosphere. But once something dies, the carbon 14 is not replenished and the amount present at the time of death starts decreasing.

So carbon dating is a process of determining the amount of carbon 14 left in a once living object then calculating the time it would take to have that much carbon 14 left.

So let’s go back to our equation for exponential decay. In order to use this equation for carbon dating, we need to know what k is for carbon 14. Now we know that the half-life of carbon 14 is 5700 years which means that given any amount of carbon 14, only half that amount will be left in 5700 years due to radioactive decay. So let’s use this fact to calculate k.

Taking this information and putting it into our equation results in

\[
{0}{.}{5}{A}_{0}\hspace{0.33em}{=}\hspace{0.33em}{A}_{0}{e}^{{-}{k}\times{5700}}
\]

So the left side shows that there is half (0.5) the initial amount and the right side shows that this occurs in 5700 years. So now, I will take the loge (abbreviated as ln) of both sides. Note that since e is the base on the right side, taking the log to that base just results in the exponent –kt. Also note that A0 appears on both sides of the equation, so we can divide both sides of the equation by A0 which makes A0 disappear:

\[
\begin{array}{l}
{{0}{.}{5}{A}_{0}\hspace{0.33em}{=}\hspace{0.33em}{A}_{0}{e}^{{-}{k}\times{5700}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{0}{.}{5}\hspace{0.33em}{=}\hspace{0.33em}{e}^{{-}{k}\times{5700}}}\\
{\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\ln{(}{0}{.}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}\ln{(}{e}^{{-}{k}\times{5700}}{)}\hspace{0.33em}{=}\hspace{0.33em}{-}{k}\times{5700}}\\
{\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{-}{0}{.}{6931}\hspace{0.33em}{=}\hspace{0.33em}{-}{k}\times{5700}}\\
{\Longrightarrow{k}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{0}{.}{6931}}{5700}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{0001216}}
\end{array}
\]

So now that we know what k is, we can use the following equation to do our carbon dating:

\[
{A}\hspace{0.33em}{=}\hspace{0.33em}{A}_{0}{e}^{{-}{0}{.}{0001216}{t}}
\]

So let’s say a fossil has 35% (0.35) of its original carbon 14 when it died. How old is the fossil?

\[
\begin{array}{l}
{{0}{.}{35}{A}_{0}\hspace{0.33em}{=}\hspace{0.33em}{A}_{0}{e}^{{-}{0}{.}{0001216}{t}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{0}{.}{35}\hspace{0.33em}{=}\hspace{0.33em}{e}^{{-}{0}{.}{0001216}{t}}}\\
{\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\ln{(}{0}{.}{35}{)}\hspace{0.33em}{=}\hspace{0.33em}\ln{(}{e}^{{-}{0}{.}{0001216}{t}}{)}\hspace{0.33em}{=}\hspace{0.33em}{-}{0}{.}{0001216}{t}}\\
{\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{-}{1}{.}{0498}\hspace{0.33em}{=}\hspace{0.33em}{-}{0}{.}{0001216}{t}}\\
{\Longrightarrow{t}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{1}{.}{0498}}{{-}{0}{.}{0001216}}\hspace{0.33em}{=}\hspace{0.33em}{8633}\hspace{0.33em}{\mathrm{years}}}
\end{array}
\]

We have a lot of birthdays to catch up on!

Logarithms, Part 4

Let’s do another example using logarithms. As seen in my last post, logarithms are useful when the unknown variable in an equation is in the exponent of some number. But the exponent can be more than just the unknown – it can be an expression with an unknown. Consider the following problem:

\[
{10}^{{3}{x}{+}{7}}\hspace{0.33em}{=}\hspace{0.33em}{125}
\]

So the first step, as seen last time, is to take the log of both sides of the equation. We then can use the property of logs that was introduced: \[ {\log}_{a}{b}^{x}\hspace{0.33em}{=}\hspace{0.33em}{x}\hspace{0.33em}{\log}_{a}{b} \]

So let’s again use the base 10 log, the log x key on your calculator:

\[
\begin{array}{l}
{{10}^{{3}{x}{+}{7}}\hspace{0.33em}{=}\hspace{0.33em}{125}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\log{(}{10}^{{3}{x}{+}{7}}{)}\hspace{0.33em}{=}\hspace{0.33em}\log{(}{125}{)}}\\
{\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{(}{3}{x}\hspace{0.33em}{+}\hspace{0.33em}{7}{)}\log{10}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{0969}}
\end{array}
\]

Now let’s stop here. The log of 125 is done on your calculator. What about the log of 10? Well that can be done on your calculator as well, but if you’ve been paying attention, you can see that the answer is 1. 1 is the exponent of 10 to make 10? = 10. So now we have a standard (non-exponential) equation:

\[
{3}{x}\hspace{0.33em}{+}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{0969}
\]

We have solved equations like this before, so without going into the detail, the solution to this is x = -1.6344. You can put this value of x in the left side of the original equation and find that it does solve it.

In my next post, I will present another property of logs and use it to solve a population problem.

Logarithms, Part 3

Finally have a little time for a post.

So we know how to solve x2 = 10 by taking the square root of both side of the equation to get x = Β±3.162… Note that taking the square root of x2 undoes or reverses the squaring of x.

But what do you do if x is in the exponent and not the base?

\[
{2}^{x}\hspace{0.33em}{=}\hspace{0.33em}{10}
\]

You can’t take the xth root since you don’t know what x is. So what to do? From my last post, you saw that log2 10 means “what is the number that I can use as the exponent of 2 so that the answer is 10”. So in the above equation, if I take the log2 of both sides, I get

\[
{\log}_{2}{2}^{x}\hspace{0.33em}{=}\hspace{0.33em}{\log}_{2}{10}
\]

The left side of this equation is doing two inverse operations on the number 2 – raising 2 to a power then taking its log. In other words, the left side can be seen as saying “what is the number that I can use as the exponent of 2 so that the answer is 2x ?”. Well the answer to that question is x. So the left side is just x and the right side is just a calculation:

\[
{\log}_{2}{2}^{x}\hspace{0.33em}{=}\hspace{0.33em}{\log}_{2}{10}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}{\log}_{2}{10}\hspace{0.33em}{=}\hspace{0.33em}{3}{.}{32192809488}
\]

Well that’s just dandy! Trouble is, without the internet, how do you find log2 10? I have not seen a calculator with a log2 x button. As mentioned in my last post, calculators usually have buttons to take logs relative to bases 10 and e. Well fortunately, there are lots of properties of logs that can help. The one we can use here is

\[
{\log}_{a}{b}^{x}\hspace{0.33em}{=}\hspace{0.33em}{x}\hspace{0.33em}{\log}_{a}{b}
\]

This means that I can take the log with respect to any base, and the x can be removed as the exponent. So for our problem, let’s take the log10 (the log x key on your calculator) of both sides and see what happens:

\[
{\log}_{10}{2}^{x}\hspace{0.33em}{=}\hspace{0.33em}{\log}_{10}{10}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}{\log}_{10}{2}\hspace{0.33em}{=}\hspace{0.33em}{1}
\]

Let’s stop here for a moment before I complete the solution. Why is the right side equal to 1? Log10 10 is saying “what power of 10 equals 10?”. The answer is 1 because 101 = 10. On the left side, I used to log property above to bring the x in front of the log. Now log10 2 is just a number. You can use the log x key on your calculator to find that log10 2 = 0.3010 to four decimal places. So now the equation becomes

\[
{x}{\log}_{10}{2}\hspace{0.33em}{=}\hspace{0.33em}{1}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{0}{.}{301}{x}\hspace{0.33em}{=}\hspace{0.33em}{1}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{0.301}\hspace{0.33em}{=}\hspace{0.33em}{3}{.}{3219}
\]

So 23.3219 = 10. In my next post on logs, I’ll do more equation solving using logs.

Logarithms, Part 2

So what is a logarithm? Let’s first see the notation, then I will explain. When taking the log (short for logarithm which I will use from now on) of a number, you need to know what base is being used. The notation for the log of x is loga x. The a is the base and is usually a specified number. so examples using this notation are log2 10, log10 25, log18 145, loge 7.34. Let’s look at these.

log2 10 is asking the question “What number can I use as the exponent of 2 so that the answer is 10?”. It turns out that 23.321928094887 = 10 so log2 10 = 3.321928094887.

log10 25 is asking the question “What number can I use as the exponent of 10 so that the answer is 25?”. Well, 101.39794 =25 so log10 25 = 1.39794.

Are you getting the picture? What about log18 145? This is asking the question “What number can I use as the exponent of 18 so that the answer is 145?”. 181.72183 = 145 so log18 145 = 1.72183.

Now let’s look at loge 7.34. This shows that the base or the number we are taking the log of does not have to be an integer. The number e, which I have talked about before, is an irrational number, but it still can be used as a base. In fact, it is probably the most used base. Since e1.99334 = 7.34, then it follows that loge 7.34 = 1.99334.

By the way, on most calculators, the log or log x key assumes that the base is 10. On most calculators as well, ln x means loge x. “ln” means “natural log”.

Now loga x and ax are inverses of each other. This means that one undoes the other. So if on your calculator, you find ln 7, then take that number and hit the ex key, you get the original 7 back. This works in reverse as well: Find e7 on your calculator, then hit the ln x key. You will again get the 7 back.

In notation-speak, this inverseness is shown as

\[
\begin{array}{l}
{{a}^{{\log}_{a}x}\hspace{0.33em}{=}\hspace{0.33em}{x}}\\
{{\log}_{a}{a}^{x}{=}\hspace{0.33em}{x}}
\end{array}
\]

In my next post, I will show how logarithms can be used to solve equations.

Logarithms, Part 1

Logarithms confuse many of my students so I thought it is time to explain these. I touched on these before on a post about inverse operations, but let’s add some more detail.

Let’s first define some terms here. Consider the expression x2. Here, x is raised to the power of 2. x is the base and 2 is the exponent, power, order, or index. Lots of different terms for the exponent – I will mostly use the term exponent. So the exponent defines what to do with the base.

Now before I talk about logarithms specifically, I want to review what various kinds of exponents mean. I have talked about this before, but these concepts should be fully understood if logarithms are to make sense to you.

Now x2 means x Γ— x. Positive integer exponents means how many times you multiply the base by itself. So in general, for a positive integer m,

xm = x Γ— x Γ— x Γ— x … where x is listed m times.

The special case of when m = 0 is defined as x0 = 1, no matter how small or how large x is. Now what about negative integers?

\[
\begin{array}{c}
{{x}^{{-}{1}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{{x}^{1}}{;}\hspace{0.33em}\hspace{0.33em}{x}^{{-}{2}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{{x}^{2}}{;}\hspace{0.33em}\hspace{0.33em}\frac{1}{{x}^{{-}{2}}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{2}}\\\
{{x}^{{-}{m}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{{x}^{m}}{;}\hspace{0.33em}\hspace{0.33em}\frac{1}{{x}^{{-}{m}}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{m}}
\end{array}
\]

So a negative exponent is the same as the positive one except it and its base is in the denominator or vice versa. You can freely move a factor that is a base and its exponent between the numerator and the denominator, as long as you change the sign of the exponent.

What about fractional exponents? Let’s start with fractions where “1” is in the numerator. The denominator in a fraction exponent refers to the root of the number. For example,

\[
{x}^{\frac{1}{2}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[2]{x}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{x}
\]

The “2” for the square root is usually assumed if it is not there. However, for other roots (like cube roots), the index must be there to indicate the kind of root it is. Other examples:

\[
{x}^{\frac{1}{3}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[3]{x}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}^{\frac{1}{6}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[6]{x}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}^{\frac{1}{n}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[n]{x}
\]

The numerator in a fractional exponent means the same as if it wasn’t in a fraction. so we can combine these two definitions for more general fractions:

\[
{x}^{\frac{2}{3}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[3]{{x}^{2}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}^{\frac{5}{6}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[6]{{x}^{5}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}^{\frac{m}{n}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[n]{{x}^{m}}
\]

Now we have not covered irrational exponents like xπœ‹. The development of these are a bit more complex so I’ll just say “use your calculator”.

Indeed, you can use your calculator to calculate a number raised to a power if it has a key labelled as “yx” or has a key with the “^” symbol on it. I will leave it to you to find out how to use these keys. If you do not have a fancy calculator, there is always the all-knowing internet.

So we have talked before on how to solve equations like x2 = 16 by taking the square root of both sides of the equation. But how do you solve 2x = 16? Notice that x is now in the exponent. That changes everything as you can’t take the xth root of a number on your calculator……………but can you?

In the next post on this topic, I’ll introduce you to logarithms then later, how they are used.

Percentages, Part 3

So now that we know what a percentage is, how is it used? Let’s look at some sample percentage problems.

Melbourne’s Silvan Reservoir has a capacity of 40,446 ML(megaliters). Currently, it is 88% full. How much water is in the reservoir? In other words, what is 88% of 40,446?

Whenever you see or interpret a problem where you need to take a percentage “of” something, equate the word “of” with “multiply”. So to take 88% of 40,446, we multiply 40, 446 by 88%. If you do this on a calculator, you need to use the decimal equivalent of 88% which is 0.88 (see my previous post). On some calculators, there is a percent key. On these, you can type 88 which the calculator will interpret as 0.88 when you use that key. Regardless of which calculator you use, when you multiply 40, 446 by 88%, you should get 35592.48 ML.

If you do not live in Melbourne, the above problem does not interest you much. As money is of interest to most everyone, let’s look at some typical money related percentage problems.

Since DavidTheMathsTutor has effectively educated the masses on how to use percentages, calculators with percentage keys are no longer in demand. So a store has discounted the normal $24.95 price of these calculators by 30%. What is the new price?

This is a two-step problem: first find what the amount of the discount is, then subtract it from the original price. The amount of the discount is 30% Γ— 24.95 = 0.3 Γ— 24.95 = $7.48 (rounded to 2 decimals as we are talking about money). So the store is taking $7.48 off each calculator. So the new price is 24.95 – 7.48 = $17.47.

On the other hand, again because of DavidThe MathsTutor, there is a big demand on fancy calculators that do all sorts of mathematical things like graph equations. So the store decides to markup the normal price of these calculators by 25% to make up for the loss of the percentage calculators. The normal price of these are $149.50 each. What is the new price?

Again, a two-step problem, but this time you are find the amount of the price increase, then add it to the original price since this is a markup. The amount of markup is 0.25 Γ— 149.50 = $37.38. So the store is increasing the price by $37.38, so the new price is 149.50 + 37.38 = $186.88.

Many times, you need to calculate the original price. For example, you are looking to buy a car. The sticker says $24,500. The salesperson says that’s a good price because they are only making a 5% profit on it. What is the cost of the car to the store? This is a reverse markup problem: what price plus 5% of that price is $24,500?

In equation form, that is the equivalent maths sentence, this is

x + (0.05 Γ— x) = 24,500.

If you remember my posts on equations, factoring, and the distributive property, this is solved by factoring the left side and then dividing both sides by the number that results from the factoring:

\[
\begin{array}{l}
{{x}\hspace{0.33em}{+}\hspace{0.33em}{(}{0}{.}{05}\hspace{0.33em}{+}\hspace{0.33em}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}{(}{0}{.}{05}\hspace{0.33em}{+}\hspace{0.33em}{1}{)}\hspace{0.33em}{=}\hspace{0.33em}{1}{.}{05}{x}\hspace{0.33em}{=}\hspace{0.33em}{24500}}\\
{\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{24500}{1.05}\hspace{0.33em}{=}\hspace{0.33em}{\$}{23333}{.}{33}}
\end{array}
\]

If you knew the cost to the store was $23,333.33 and knew that they marked up that cost by 5%, if you calculated the new price as we did before, you would get $24,500.

Sometimes you need to calculate the actual percentage. What if your salary went up from 88,000 to 99,000? What percentage pay rise is this (so you can brag to your friends)?

The amount of pay increased by 99,000 – 88,000 = 11,000, so we want to know what percentage of 88,000 is 11,000. Note that you always work with the original price or amount when working out a percentage. So the equivalent maths sentence is

88,000 Γ— x% = 11,000

Dividing both sides by 88,000 gives

\[
{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{11000}{88000}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{125}\hspace{0.33em}{=}\hspace{0.33em}{12}{.}{5}{\%}
\]

You must be very good at your job!

Factors of Polynomials, Part 4

So we are on a journey of how to factor a polynomial like

\[
{x}^{3}\hspace{0.33em}{-}\hspace{0.33em}{6}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{x}\hspace{0.33em}{+}\hspace{0.33em}{30}
\]

and the motivation to do this is so we can easily solve equations where the polynomial equals 0. For example, if we have

\[
{(}{x}{-}{7}{)(}{x}{+}{3}{)(}{x}{-}{2}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

where the left side is a factored form of a polynomial, you can quickly see that x = 7, -3 or 2 will solve the equation since these numbers make one of the factors 0.

The last post reviewed some of the tools you can use if you have a quadratic (highest power of x is 2) polynomial. But what can you do if you have a polynomial like the one at he beginning of this post? Well there are two theorems regarding polynomials that will help here: The Remainder Theorem and the Rational Root Theorem.

The modified Rational Root Theorem says that if a polynomial has an integer root (that is a value of x that makes the polynomial equal to 0), then that value of x must be an integer root of the constant term (the number without any x‘s). This is a modified version of the Rational Root Theorem because I am restricting the coefficient in front of the highest power of x to be 1.

So for our polynomial, the constant term is 30. So the possible roots are Β±1, Β±2, Β±3, Β±5, Β±6, Β±10, or Β±15 as these are all factors of 30. So how do we check if any of these numbers are a root? Along comes an application of the Remainder Theorem: If a is a root of a polynomial (that is (xa) is a factor of the polynomial), then the polynomial evaluated at a equals 0. This is really a specific application of the more general Remainder theorem, but this is all we need.

What this means is that if I want to test if a number is a root of a polynomial, all I have to do is evaluate (replace the x‘s) with that number and see if I get 0. If I do, the the number is a root, if not, then it’s not.

So let’s check if 1 is a root, that is check if (x – 1) is a factor:

\[
{1}^{3}\hspace{0.33em}{-}\hspace{0.33em}{6}{(}{1}{)}^{2}\hspace{0.33em}{-}\hspace{0.33em}{1}\hspace{0.33em}{+}\hspace{0.33em}{30}\hspace{0.33em}{=}\hspace{0.33em}{24}
\]

So (x – 1) is not a factor. Let’s try -2:

\[
{(}{-}{2}{)}^{3}\hspace{0.33em}{-}\hspace{0.33em}{6}{(}{-}{2}{)}^{2}\hspace{0.33em}{-}\hspace{0.33em}{(}{-}{2}{)}\hspace{0.33em}{+}\hspace{0.33em}{30}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

Great! That means (x + 2) is a factor. Now from here we can do two things: either divide our original polynomial by the newly discovered factor (x + 2) or we can continue using the Remainder Theorem to test the other integer roots.

Polynomial division is not hard but It’s just as easy to continue using the Remainder Theorem. If you check x = 3 and x = 5, you will see that the polynomial does equal 0. So the compete factorisation of our polynomial is:

\[
{x}^{3}\hspace{0.33em}{-}\hspace{0.33em}{6}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{x}\hspace{0.33em}{+}\hspace{0.33em}{30}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}{+}{2}{)(}{x}{-}{3}{)(}{x}{-}{5}{)}
\]

Don’t you feel mathematically powerful!!

Factors of Polynomials, Part 3

As the order of the polynomial (the highest power of x) increases, it usually gets harder to factor. In my last post on this topic, I will cover a way to reduce the order by one for each iteration of the process. If you can get the polynomial to a degree 2, there are many ways to factor these.

A polynomial of degree 2 is called a quadratic. I covered factoring quadratics or solving quadratic equations (equations where the quadratic is set equal to 0) in several posts before. Please review these but I will just remind you of them here.

A quadratic is a polynomial of the form \[{ax}^{2}\hspace{0.33em}{+}\hspace{0.33em}{bx}\hspace{0.33em}{+}\hspace{0.33em}{c}\hspace{0.33em}\]

whereΒ a, b, andΒ cΒ are some numbers.

Now for this set of posts, I am restricting a to be 1. So we would like to factor the quadratic to look like \[{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}\]

Basically, the method is to do a reverse distributive property (please see my posts on this). Let’s do an example. Let’s factor \[{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{24}\hspace{0.33em}\]

We need to find two numbers, aΒ andΒ b, so that they multiply to equal -24 and add or subtract to equal +2, the coefficient in front of theΒ x. 8 and 3 do not work as they do not add to equal 2. However, 6 and 4 look like contenders. The signs of 6 and 4 must be such that they add to equal +2 but multiply to equal -24. Looks like +6 and -4 work so \[{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}{24}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{6}{)(}{x}\hspace{0.33em}{-}\hspace{0.33em}{4}{)}\hspace{0.33em}\]

Another method you can use is to find the zeroes of the quadratic directly instead of factoring. This method is the quadratic formula. Please see my prior post on this.

The quadratic formula is \[{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{b}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{{b}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{ac}}}{2a}\]

where a, b, andΒ cΒ are the coefficients in the general form of a quadratic.

From the example I just factored, we can see that x = -6 and x = 4 are zeroes of the quadratic. I could find these directly using the quadratic formula:

\[
{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{2}\pm\sqrt{{2}^{2}{-}{4}{(}{1}{)(}{-}{24}{)}}}{2(1)}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{2}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{100}}{2}\hspace{0.33em}{=}\hspace{0.33em}{4}{,}\hspace{0.33em}{-}{6}
\]

So looks like we have a few tools available to factor quadratics. But what can we do if the order of the polynomial is higher than 2? I will cover a method to do this in my next post.