In my previous post, I introduced the concept of logarithms to you. I explained how logarithms and exponents are connected, and then showed you a quick trick to help you remember how to convert between the two. Now in this post, I’m going to go a little bit deeper and explain a few rules of logarithms to help you actually do math with them. Considering how you now already know that logs and exponents are related, it should come as no surprise that, just like there is a set of exponent rules, there is also a set of logarithmic rules.

As you look at these logarithm rules, keep in mind that by convention, if you write logs without the subscript number to indicate their base, it is assumed that you are dealing in base–10. For simplicity, this is the convention that I am going to use in this post, though these rules certainly apply when dealing with logs of other bases.

With that intro out of the way, let’s get to it.

The first law of logarithms is the product rule. If you are familiar with the product rule of exponents, then this logarithm law should be a piece of cake for you. Where the exponent rule says that when multiplying exponential expressions with the same base, you simply add the exponents, this same thing applies when multiplying logarithms of the same base. Therefore, the rule states that the logarithm of a product is equal to the sum of the logarithms.

This rule is very commonly used, and it is important to recognize that you can use it in either direction. That is, the logarithm of a product converts to a sum of logarithms, and vice versa.

The next logarithmic law is the quotient rule. Again, this law can easily be derived by applying your knowledge of the exponent quotient rule (though I will leave that for you as an exercise). However, it does appear to look different. This rule states that the logarithm of a quotient is equal to the logarithm of the numerator minus the logarithm of the denominator. Sounds like a mouthful, but the expression is probably much simpler to understand.

Again, watch for opportunities to use this relationship in either direction!

The third law of logs is the power rule. This one is surprisingly simple to remember, and again can be found by manipulating exponent and logarithm laws. Quite simply, this law says that when you have a logarithm of an exponential expression, the exponent can be “brought out” of the log and used as a coefficient for the log.

The last of the rules of logarithms that I’m going to discuss here today is called the base change rule. Recall that I stated above that all of my examples that I’ve used so far in this post use the convention of an assumed base–10. If I wanted to change my expression to utilize a different base, this rule helps us to do that. So then, if I have my log in some base of a number, and I want to express this in terms of a different base, I simply take the log in my new base of the original number and divide that by the log in my new base of the original base. Sounds wordy, but again, a picture is worth a thousand words:

Here, my original base is B, and my new base that I want to express things in is X.

That is all I have to say about the rules of logarithms in this short introduction to them. They are fairly straight forward themselves, though can be used in very complex equations. I will try to do a separate post soon outlining some examples of all of these rules, though I do think that the general forms that I’ve noted above are pretty self-explanatory.

If you are interested in learning more about logarithms, there is a much more thorough summary of logarithms at the Learning and Teaching Math blog, which I highly recommend (for this and other math topics!)

Pythagorean Identities in trigonometry will show up very frequently and can be very useful.  I will explain how Pythagorean Identities get their name, how you can derive them, and how you can remember them.  First, it would be a good idea for you to be able to understand the basic trig functions sine, cosine, and tangent.  Once you are familiar with these trig equations, the algebra that we will apply to them will allow us to derive the Pythagorean Identities. I have prepared other posts on this site that are dedicated to sine, cosine, and tangent that may be useful to review.

The Pythagorean Identities get their name because they are based on the famous Theorem of Pythagoras.  You are very likely already familiar with it. (On a side note, here are some interesting facts about the Theorem of Pythagoras.) Simply, for a right angle triangle, it says “the square of the hypotenuse is the sum of the squares of the other two sides.”  Mathematically, you have seen this represented as:

a² + b² = c², where a and b are sides and c is the hypotenuse. 

Now, I will show you how to derive these special trig identities, using this theorem as our starting point.  To do this, we need to start with a right triangle, created by the radius of a unit circle and the axis:

We can say that the right triangle formed by dropping a line from the point that the radius touches the circle (anywhere in quadrant I is sufficient for this demonstration) down to the axis has a base of x units long and y units high. (The actual numbers are not important, but they will depend on the specific angle, if you did need to calculate them for whatever reason. You don’t here.)  The radius in a unit circle, by definition, is 1.  Now, let’s apply the definitions of sine and cosine to our triangle.  Recall:

sin(ɵ) = opposite / hypotenuse = y / 1 = y

cos(ɵ) = adjacent / hypotenuse = x / 1 = x

So, now we can relabel our diagram by substituting in these basic trig identities.

With the triangle now correctly labeled for our derivation, we can apply the Theorem of Pythagoras to arrive at one of the Pythagorean Identities. Since a² + b² = c², we can therefore equate the sides of our triangle to these terms to give us our first of the trig Pythagorean Identities:

sin²(ɵ) + cos²(ɵ) = 1

If you have followed along up till now and understood everything I’ve done, then you are well on your way to remembering this trigonometric identity.  If you can remember how to derive it, you don’t even have to memorize it (though it always helps!)  For the next Pythagorean Identity, you start with this first identity, and you apply some basic algebra and trigonometry to it to derive the second and third identities.  Recall the definitions of secant, cosecant, and cotangent:

sec(ɵ) = 1 / cos(ɵ)

csc(ɵ) = 1 / sin(ɵ)

cot(ɵ) = 1 / tan(ɵ) = cos(ɵ) / sin(ɵ)

With those inverse trig functions in mind, let’s take the first Pythagorean Identity and divide all of its terms by cos²(ɵ).  That gives you: 

1 / cos²(ɵ) = sin²(ɵ) / cos²(ɵ) + cos²(ɵ) / cos²(ɵ)

sec²(ɵ) = tan²(ɵ) + 1

And this is the second Pythagorean Identity!  Using the same strategy we just used to derive that one, go back to the first one and divide everything by sin²(ɵ), to arrive at the third Pythagorean Identity!

csc²(ɵ) = 1 + cot²(ɵ)

I hope that from this tutorial, you now understand how these identities get their name, how you can derive them, and how to use this knowledge to help you to memorize or recall them.  Using the fundamental trigonometry identities and trig relations, it is easy to come up with more advanced trigonometric formulas.  If you need to refer back to this Pythagorean Identities list, please bookmark this page and come back again.  Also, be sure to follow me on Facebook/Twitter/Google+ (@TheNumerist).  Thanks.

(This is an old post from my previous math site, In Mathematics, copied here to consolidate all my math pages.)

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In this post, I’m going to explain a very frequently requested topic – how to convert your equation of a line from point-slope form to standard form. Sounds easy, right? Well, it isn’t difficult at all, provided that you understand the terminology and know what you’re doing. Follow along and hopefully all will become clear!

You are familiar with the general form of y=mx+b (also known as slope-intercept form), and you know that this equation tells you all that is necessary to actually graph this line – namely, the slope and y-intercept. However, what about if you have a section of your line up in quadrant I at the ordered pair coordinate of (150332, 23098)? The y-intercept on this graph doesn’t seem terribly useful way over here at this distant point! In this case, it’s probably more appropriate to use the point-slope form for your equation of a line. I need to explain this form first, before going on to show you how to convert from point-slope form to standard form equations.

In the most simple explanation that I know of, you can very easily derive the point-slope form from a very well known concept: the slope formula! Recall that slope is equal to rise over run. But what does that mean, in terms of mathematical symbols. Well, as I explained already in a previous post, this refers to the difference in the y values between two points, divided by the difference of the x values between those same two points. In formula form, you get something like this to define slope:

Now, to arrive at the point-slope form, all you need to do is a very simple rearrangement, as follows. Then, let the y₂ and x₂ just be x and y, and you are left with what you need to know:

Hopefully, you can see the manipulation that I did there. I simply multiplied both sides by the denominator, and then switched sides so that you could see the more conventional form of this equation of a line. The 1’s and 2’s aren’t particularly important – here, the y₁ and x₁ terms are simply referring to a specific point, whereas x and y refers to any point.

To actually use this equation, you have a few ways. In one way, you can substitute in the m value and a given coordinate that is on your line for the y₁ and x₁ terms, and then go from there to simplify or solve for another point. Secondly, you can use two separate points to calculate the slope (remember, this is essentially just a rearranged slope formula!) Either way, this form of the equation of a line is incredibly useful and handy to know. And thankfully, being able to derive it easily from the slope formula gives you an easy way to come up with it if you can’t seem to remember it exactly when you need it the most (on exams!).

So, now that you know what point-slope is, let me refer you back to my previous post about standard form graphing equations – because, now I’m going to explain to you how to convert from point-slope to standard form. This isn’t a terribly complicated process, though it is extremely important to get right, because when done correctly, both forms mathematically represent the same line on a graph. Though, if you make an error, you will likely wind up with a different line altogether. It is important to pay attention to what form of the equation of a line you are being asked to provide, and then it’s just a matter of doing some of these steps!

Point-Slope Form to Standard Form

Example: Express the following equation in standard form, and state the values for A, B, and C.

As a first point, I want you to realize that this example is very explicitly provided in point-slope form – to the letter! It won’t always be so! In any case, here is the basic strategy of what you want to do: get all of the x’s and y’s together on one side, and get the constants (i.e. no variables) over to the other side. Then, it’s just a matter of combining like terms and simplifying things wherever possible. Probably the most important thing to remember here is that you need to multiply what’s inside the brackets by the constant on the outside! This is far too easy a step to miss, but will completely mess you up!

Hopefully you can follow along with those steps! All it is really is a rearrangement of the terms, grouping the x’s and y’s together, and the constants alone. When you get it into the final form as I have shown, it is easy to simply read off the values for A (the coefficient in front of the x), B (the coefficient in front of the y), and C (the constant with no variable attached to it). In this case, A is 2, B is -1, and C is -10. Remember, no number in front of the y means a 1 is assumed, and since the standard form has a +, in this case, the minus means there is a -1.

Try another one, a bit harder this time?

Example: Express the following equation in standard form, and state the values for A, B, and C.

In this case, note that it isn’t immediately in point-slope form – I’ve reversed the left side terms. Of course, it’s a simple matter of just rearranging these, like so:

There, now that’s more appropriate. Next, we just follow the same steps that we did above: multiply through the brackets, and then group the x’s and y’s and isolate the constants. Easy, right? Let’s see what we get.

I did all of the adding and subtracting on one line this time, but I did the same steps as I outline above, and as you can see, I have a final answer expressed in standard form! If you were to stop here, and say that A is 2/3, B is 1, and C is 7, you would most definitely be correct. However, there is a convention that many teachers and professors follow, and that is to remove everything from the denominator, wherever possible. In other words, teachers don’t like fractions! So, how do we get rid of our fraction? You probably have already figured out where I’m going with this – you simply have to multiply everything on both side by 3, the denominator, to cancel it out. Doing so, you wind up with this final standard form graphing equation:

In this case, A is 2, y is 3, and C is 21. Another note – these coefficients are different from those we originally got, but the underlying math is all the same still. You can take both forms of our answers, create a table of values for each, and manually plot out the lines to prove that these indeed are the same lines, even though the equations look a bit different. You will probably agree that this version of the equation of the line just looks a lot nicer.

So, hopefully those few examples have properly explained to you the steps to consider when you have to convert point-slope form to standard form graphing equations. It’s not as difficult as it sounds, you just have to remember the points I’ve described in this post. In the next post, I’ll expand this concept to explain how slope-intercept form fits into all of this. Eventually, you won’t even recognize what form you are actually working with. You will just recognize what you need to do with the numbers to get the information that you need to solve your problem.

Thanks for reading this rather lengthy post! Please remember to subscribe or click on one of the Follow buttons on the right side of this page! I appreciate the support! And don’t forget, comments are always welcome if you need more explanations!

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