## Functional Notation, Part 2

Last time we saw that we can replace y in an equation with f(x) when y is alone on the left side of an equation:

y = f(x) = 3x² – 5x + 1

The above is an example of the function definition. Once defined, you replace all the x‘s on the right side with whatever is in the brackets on the left side, even if it is not a number. For example,

f(2) = 3(2)² – 5(2) + 1 = 3
f(a) = 3a² – 5a + 1

Even if the thing in the brackets is another expression, for example, an expression that is used in calculus a lot is x + h:

f(x+h) = 3(x + h)² – 5(x + h) + 1

And you can even use another function of x inside the brackets of another function. Like x, the letter f is used in the first instance for a function, but if other functions need to be defined as well, other letters are used:

f(x) = 3x² – 5x + 1
g(x) = x² – 7
f[g(x)] = f(x² – 7) = 3(x² – 7)² – 5(x² – 7) + 1
g[f(x)] = g(3x² – 5x + 1) = (3x² – 5x + 1)² – 7

The domain of a function is all the valid values of x that can be used. Many times, the domain of a function (like f(x) and g(x) above) is just any real number. But there are functions where you cannot use just any number. For example, consider

${f}{(}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{{x}{-}{2}}$

There is one value of x you cannot use. That value is 2 because that will make the denominator 0, and as you know, this will bring the maths police to your door. So the domain of this function is all real numbers except for 0.

Now consider

${f}{(}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{{x}{-}{2}}$

Another illegal operation is taking the square root of a negative number. The requirement for this function is that x – 2 has to be 0 or greater. For this to be true, x must be greater than or equal to 2. The phrase ” greater than or equal to” can be replaced by the maths symbol ≥. So the domain of this function is x ≥ 2.

There are other reasons why the domain of a function is restricted, but the most common things to look for is dividing by 0 or taking the square root (or any even root) of a negative number.

Posted on Categories Algebra, Pre-VCE

## Functional Notation, Part 1

Any maths beyond algebra relies on something called functional notation. I used this in one of my posts on Newton’s Laws but more needs to be said if you are to be comfortable with it.

We have looked at many equations to date. Most involved x and y. For example,

y = 3x² – 5x + 1

This will plot as a parabola on an xy coordinate system. The plot is a picture of all (x, y) pairs of numbers that, when substituted in the above equation, will result in a true statement. For example, (0, 1) would be a point on that parabola because 1 = 3(0)² – 5(0) + 1 = 1.

Most of the equations we have looked at have y on the left side and all the other things with x and stand-alone numbers are on the right side. In this form, it is easy to choose a number to replace x with, then do the maths with that number on the right side to find the corresponding y to make the equation true. For example, let’s choose “1” in place of x. The corresponding y will be

y = 3x² – 5x + 1 = 3(1)² – 5(1) + 1 = 3 – 5 + 1 = -1

So (1, -1) is also a point on this equation’s plot.

This is frequently done: choose a value for x, then replace x with that value and do the maths on the right side and find the corresponding y. Notice that we are free to choose a value for x, but once we do, the value for the corresponding y is fixed. For that reason, x is called the independent variable and y is called the dependent variable: y depends on the x we choose.

If y depends on the x we choose, then another way to say this is that “y is a function of x“. The new functional notation makes use of this by replacing y with f(x), read “function of x“. So the functional form of the equation above is

f(x) = 3x² – 5x + 1

This will plot exactly the same but we would replace the y label on the vertical axis with f(x):

So now it is easy to ask the question “What is the value of the function if x = 0″ by just replacing the “x” in f(x) with “0”, that is, f(0):

f(0) = 3(0)² – 5(0) + 1 = 1

So, f(0) = 1. We saw above that f(1) = -1. So now you will see the general coordinate as (x, f(x)) instead of (x, y). This is just a difference in notation – the plot stays the same.

There are some properties of functions and a few more definitions that will be explored next time.

Posted on Categories Algebra, Pre-VCE

## System of Equations, Part 2

So last time I solved a system of two equations using the substitution method where the information from one equation is inserted into the other equation. This is the method of choice if it is easy to solve for one of the unknowns. However, the example that I used also lends itself well to the other method: Elimination.

The elimination method, like the substitution method, uses the two equations to generate one equation with one unknown which can be solved. The system I worked on last time was:

x + y = 108
xy = 38

The elimination method is simply adding the two equations together with the goal of eliminating one of the unknowns. So let’s add these two equations:

x + y = 108
xy = 38
2x = 146

Notice that I now have an equation with one unknown. Dividing both sides of this new equation by 2 gives:

2x/2 = 146/2 ⟹ x = 73

the same answer as before. Now use this value for x in one of the original equations. Let’s use the first one:

x + y = 108 ⟹ y = 108 – x = 108 – 73 = 35

So we get (thankfully) the same solution as before. But this example is rather contrived in that the y‘s were conveniently of opposite signs in the given equations. So let’s consider the following system:

2x + 3y = 51
3x + 2y = 49

Adding these two equations together will just give us another equation with two unknowns. But just like we do with single variable equations, we can modify one or both of these. Let me take the first equation, and multiply it by – 2. Why I am doing this will soon be revealed:

-2(2x + 3y) = 51(-2) ⟹ -4x -6y = -102

I will now multiply the second equation by 3:

3(3x + 2y) = 49(3) ⟹ 9x + 6y = 147

Now let’s repeat the system, replacing the equations with the new ones:

-4x -6y = -102
9x + 6y = 147

Notice that if I now add these equations, the y variable will disappear:

-4x -6y = -102
9x + 6y = 147
5x = 45 ⟹ x = 45/5 = 9

I will now substitute this partial solution into the original second equation:

3x + 2y = 49 ⟹ 3(9) + 2y = 49 ⟹ 27 + 2y = 49 ⟹ 2y = 49 – 27
⟹ 2y = 22 ⟹ y = 11

So x = 9 and y = 11 solves both of these equations.

Both methods, substitution and elimination, can either be used to solve a system of equations, but one method may be less work than the other.

Posted on Categories Algebra, Pre-VCE

## System of Equations, Part 1

The equations that I have solved so far, have been equations with one unknown (variable). Suppose we have two unknowns in a problem and wish to solve for both. There is a rule in maths that you have to have the same number of equations as the number of unknowns if you are to solve for all unknowns. These equations need to be independent. what do I mean by that?

Suppose I need to find two numbers that add up to 108. In equation form, I want to find x and y such that x + y = 108. So I need another piece of information (equation) if I need a specific solution. Suppose you say “I can get another equation by just multiplying both sides of the given equation by 2: 2x + 2y = 216.” The problem here is that this second equation depends on the first equation. This is what I meant by “independent”. The second equation must be completely new information.

So let’s say that these two numbers must also have a difference of 38 : xy = 38. This is a completely new (independent) requirement. So we now have the two equations:

x + y = 108
xy = 38

So what two numbers satisfy these two requirements? There are two main methods for solving this: Substitution and Elimination. Actually, there is a third method which uses matrices, but I will explain what a matrix is and how to use it in a subsequent post. Let’s first talk about the substitution method.

The substitution method is basically inserting information from one of the equations into the other equation. Let’s solve the second equation for x. That is, use algebra to get x on one side and everything else on the other side. If I add y to both sides of the second equation, I get

xy + y = 38 + yx = 38 + y

Now put this definition of x into the first equation. That is, replace x in the first equation with what x is equal to from the second equation:

x + y = 108 ⟹ 38 + y + y = 108 ⟹ 38 + 2y =108

Notice that this new equation, which combines the information of the two equations into one, is a single equation with a single unknown. This can now be solved with the techniques we have seen before:

38 + 2y =108 ⟹ 2y = 108 – 38 = 70 ⟹ y = 70/2 = 35

So y is 35. You can now use this solution into any of the equations involving x and y, then solve for x. Let’s use x = 38 + y:

x = 38 + y = 38 + 35 = 73

So the two numbers are 35 and 73. They both add up to 108 and subtract to equal 38.

I’ll start the next post doing another example.

Posted on Categories Algebra, Pre-VCE

## Newton’s Clock, Part 4

So here is the last of this series explaining the expressions on Newton’s clock:

We are now up to 8. Let’s look at

$\mathop{\prod}\limits_{{k}{=}{0}}\limits^{1}{{(}{2}{k}{+}{2}{)}}$

This is another excellent example of how concise the words in maths can be. The symbol “𝚷” is the capital version of 𝜋 which corresponds to the english “P”. The “P” here stands for “product” which is the result of multiplying two or more numbers. The expression on the clock means : “Take the expression 2k + 2 and successively replace the k with the number at the bottom of the 𝚷 symbol (in this case “0”), evaluate the expression to get a number, and increment k by 1 and repeat until you reach the number at the top of the 𝚷 symbol (in this case “1”). Then multiply all these numbers together.”

You can see why maths expressions are much more concise than English. So to evaluate this expression, we first replace the k with 0, then work out 2(0) + 2. This equals 2. Now increment the k by 1 to get 1, then work out 2(1) + 2. This equals 4. Since k is now at the number at the top of 𝚷, we are done increasing k. Now multiply these numbers together. 2 × 4 = 8 which is the correct number at this position on the clock.

Most of you now know what

$\sqrt{81}$

means. It means “what number multiplied by itself equals 81?” The answer, of course is 9 as 9 × 9 = 81.

The next hour is

${\log}_{2}1024$

The basics of this expression have already been explained for position “2” on the clock. This expression is asking the question “what does the exponent of 2 have to be so that 2x = 1024?”. Hopefully, the answer is “10” and it is because 210 = 1024.

Now let’s look at B16. Remember when I explained position “7” on the clock: 01112? That was a number in the base 2 system of counting. Another common base used with computers is the base 16 counting system. We are familiar with the base 10 counting system that has 10 symbols used to count with: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The base 16 system needs 16 symbols. So what is used after 9 is reached? Well, we resort to the letters of the alphabet. The numbers up to 9 in base 16 correspond to the same numbers in base 10. The next number in base 16 is “A” which corresponds to 10 in base 10 and the next number is “B” which is 11 in base 10. So B16 = 11.

Finally, the last expression:

$\mathop{\sum}\limits_{{i}{=}{1}}\limits^{3}{{(}{3}{i}{-}{2}{)}}$

The Σ symbol is the Greek capital “sigma” and corresponds to the english “S”. This letter stands for “Sum” which is the addition of two or more numbers. This expression is just like the one in position “8” on the clock except that you add the resulting numbers together instead of multiplying them. So starting at i = 1, 3(1) – 2 = 1, 3(2) -2 = 4, 3(3) – 2 = 7, and we are done as i now equals the numbers on top of the Σ. So now add these numbers together: 1 + 4 + 7 = 12. It is now high noon and that completes Newton’s clock.

Posted on Categories Algebra, Pre-VCE

## 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 10² = 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 2² – 1² = 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).

Posted on Categories Algebra, Pre-VCE

## 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, and uses radians, trigonometric (or circular) functions, exponentials, and Euler’s identity. The two parts of this expression are e𝜋i and cos 𝜋. Let me first talk about cos 𝜋.

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 radians. If your calculator is in radians mode and if you have ‘𝜋’ button, taking the cosine of 𝜋 will show -1. If you do not have a 𝜋 button, take the cosine of 3.14179 and you will get an answer approximately equal to -1. So, cos 𝜋 = -1.

Now the e𝜋i part requires a bit more of an explanation. The numbers e and 𝜋 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 is exactly 𝜋. 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 e𝜋i where i is in the exponent? There is an identity (a maths rule) called Euler’s identity that explains what e𝜋i is equal to:

e𝜋i + 1 = 0

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

So if e𝜋i = -1 and cos 𝜋 = -1, then

e𝜋icos 𝜋 = (-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.

Posted on Categories Algebra, Pre-VCE

## 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!

Posted on Categories Algebra, Pre-VCE

## 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.

Posted on Categories Algebra, Pre-VCE

## 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.

Posted on Categories Algebra, Pre-VCE