## 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, 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 = ±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 × x. Positive integer exponents means how many times you multiply the base by itself. So in general, for a positive integer m,

xm = x × x × x × 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. 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.