here are the top 10 most important things to know about logarithms must know number one what is a logarithm a logarithmic function has two main components it has a base and it has an argument this subscript B is what we call the base of the logarithm and this x is the argument of the logarithm and when reading this we pronounce this as log base B of X now what does this function do well a logarithm is a function that gives the exponent to which a fixed number the base must be raised to to produce a given number the argument now let me give you a couple logarithmic Expressions to help you understand what this function does let's say we have log base 3 of 9 what this function means it means what exponent goes on three to make it equal to 9 3 to what exponent equals 9 well 3^ SAR is 9 so the answer to this logarithm is 2 because 3 raed to a^ of 2 is equal to 9 let's try another one how about log base 2 of 8 that just means what exponent goes on two to equal 8 well I know 2^ 3 = 8 so the value of that logarithm is 3 and let's do one more what if we have log base 2 of 1/8 since we're trying to figure out what exponent goes on two to make it equal to 1 over8 rewriting 1 over8 as a power of two will make the answer obvious so I could rewrite this as log base 2 of 1 over I can change that 8 to a 2 cubed and now I can take this power of 2 cubed and move it up to the numerator if I change the sign of the exponent and now it should be obvious the answer to this expression is -3 right this is the exponent that goes on the base of the logarithm to make it equal to the argument now before I move on you should know about two very special logarithms the common logarithm and the natural logarithm the common logarithm is a logarithm that has a base of 10 so if you have log base 10 of X you don't actually have to write the base you could just write log of x and if you don't see a base it's impli that the base is 10 the natural logarithm is actually a logarithm that has a base of e which is Oiler number that approximately 2.7 value if you have a log where the base is e you can communicate that instead of writing log base e you can just write Ln which stands for the natural logarithm of X so if you see either of these Expressions you have to know they're actually equal to those logarithms if if you were asked to evaluate log of 100 you'd have to know that even though you don't see a base on that logarithm the base there is 10 because it's the common logarithm so this expression means what exponent goes on 10 to make it equal 100 well 10^ SAR is 100 so the value of that logarithm is two must know number two converting between exponential and logarithmic equations let's start by looking at if we have a logarithmic equation how could you convert that to an equivalent exponential equation well logarithmic equation typically looks like y equal log base B of X and remember what does this equation mean so since Y is the answer to this logarithm I know that Y is the exponent that goes on B to equal x so using that logic I could rewrite this in exponential form as B to the^ of Y is equal to X so let's practice that with a couple examples what if we have the logarithmic equation 5 equal log base 2 of 32 well what does this equation mean it means that 2 to the^ 5 is equal to 32 so in exponential form this equation would look like 2 ^ 5 = 32 what about the equation 2 = log base 7 of let's use a variable this time x rewriting this in exponential IAL form well I know that this equation tells me that 7^ 2 is equal to X so an exponential form that's how it would look 7 ^ 2 = x so now that you can go in this direction from logarithmic to exponential form let's make sure you can go the other way from an exponential equation to a logarithmic equ to an equivalent logarithmic equation so let's say we had an exponential equation B to the^ of M = n and I wanted to write that as an equivalent logarithmic equation well I know the exponent is the answer to a logarithmic expression so I'll set my equation equal to the exponent so m equals and since m is the exponent on B I'll make B the base of my logarithm the base of the power becomes the base of the logarithm and since B to the power of M is equal to N I know that n is the argument of my logarithm let's say we have the exponential equation 4 cubed = 64 the equivalent logarithmic equation the exponent is the answer to the log so 3 equals log the base of the power becomes the base of the log so it becomes log base 4 of 64 both of these Expressions communicate the exact same information that 4 to the^ 3 equals 64 must know number three the graph of a log function this is going to be very important to understand so that you understand the restrictions of a log function let's start by picking a log function that we want to graph how about y equal log base 2 of X and when picking points for a log function it's actually easier if we pick values for y and then calculate their corresponding values of X and that's because if I rewrite this logarithmic equation in exponential form I know that this equation tells me that Y is the exponent that goes on two to equal x so I could rewrite it as 2 to the^ of y equal x so I'll pick some values for y I'll just pick values from -2 to 2 and then I can calculate the value of x just by taking those y's subbing them into this exponential equation and get the corresponding value of x let me plot these points so we can get a general idea of what the function looks like and then I'll tell you a couple of its properties now before I connect these points notice this function has a vertical ASM toote at xal 0 so I'll draw that in on our function here notice as this function approaches the vertical ASM toote from the right side of it the value of the function starts to go towards netive Infinity that's why there's a vertical ASM toote at xals 0 so let me actually make a comment about the domain of this function the domain just means what values can X be in the equation or graphically horizontally where does this function exist well the function only exists at X values bigger than zero so that means this expression will only have an answer when the argument is bigger than zero so for the domain I could say x could be any real number given X is greater than Z before I move on let me tell you about a couple general rules for log base B of X the first rule I want to tell you about is about the argument of the logarithm when we looked at this function log base 2 of X we saw it only existed when X was bigger than zero so we now know that the argument of a logarithm always has to be bigger than zero so I could say that X which is the argument of the log has to be bigger than zero and it's easiest to understand that when I give you a counter example so what if I gave you a log expression where the argument was not bigger than zero what if I gave you the expression log base 5 of -25 and asks you to find the value well that would mean what exponent can I put on five to make it equal to -25 it doesn't matter what exponent you put on a positive number it's never going to become negative and the other property you have to know is about the base of the logarithm notice when I chose the base of this function we analyzed I made it a positive number and that always has to be the case the base of a logarithm has to be bigger than zero and once again if I give you a counter example how about log base -4 of 2 that would mean what exponent can I put on -4 to make it equal two well that exponent doesn't exist and there's another condition actually not only does the base have to be bigger than zero it also cannot be one an example of why that can't happen well what if we had log base 1 of two that means what exponent goes on one to equal to well it doesn't matter what exponent you put on one it's always going to be equal to one there's no way it can get equal to two so for that reason a logarithm cannot have a base of one must know number four the power rule of logarithms here's the first main rule that you can use when working with a logarithmic expression if you have log base B of A power so of M to the power of n there's a log rule that says that you can take the exponent of the argument and rewrite it as the coefficient of the logarithm so this would be equal to n * log base B of M now this rule can only be used if the base of the log is bigger than zero and not one and also if m is bigger than zero right because m is the argument of the log we know the argument of a logarithm has to be bigger than zero now let me show you how we could use this rule let's say you had the expression log base 3 of 9 ^ 4 there's a few ways we could think of evaluating this but it might be easy if you notice that the argument is a power so you can take the exponent of that argument and rewrite it as the coefficient of the logarithm so this would be equal to 4 * log base 3 of 9 and then log base 3 of 9 well that just means what exponent goes on 3 to equal 9 3^ 2 is 9 so that logarithmic expression is equal to 2 so I have 4 * 2 which is equal to 8 let me move this and show you another example let's say we had log base 2 of < TK 8 well remember that square root symbol can be written as a rational exponent it can be written as 8 ^ of 1 / 2 and now the power rule of logarithms tells me I can take that exponent of the argument and rewrite it as the co efficient of the logarithm and log base 2 of 8 just means what exponent goes on two to make it equal to 8 2 to the power of 3 is 8 so I know that logarithmic expression is equal to 3 and a half * 3 is 1.5 or 3 over2 this power rule of logarithms is also useful for solving different types of equations let me give you a couple equations that we could use this rule to help solve the first one is the equation 2 equals log of 3 ^ x since this x is the exponent of the argument of a logarithm I can write it as the coefficient of the logarithm and then I could just divide both sides by log 3 and if I use my calculator to get an approximate value for that it's about 4.19 a second example where the power rule of logarithms would come in handy would be the equation 100 equal 2 to the^ of Y now there's a couple ways we could solve this equation but one way would be to take the logarithm of both sides and then to use the power rule of logarithms to take that exponent of the argument and write it as the coefficient of the log and then also notice log 100 well that's the common logarithm the base of the log even though we don't see anything there is 10 so that means 10 to what exponent is 100 well 10^ squ is 100 so the answer to that log is 2 so I have 2 = y * log 2 I can divide both sides by log 2 to get Y = 2 / log 2 and then using a calculator I could get an approximate value for that of 6.64 must know number five the product and quotient rules of logarithms let's start with the product rule of logarithms the product rule of logarithms tells me that if you are adding two logarithms that have the exact same base so log Bas B of M plus log base B of let's say n you could rewrite the sum of those two logarithms as a single logarithm if you keep that same base B but multiply the arguments together so M * n Now when using this rule you have to be careful it can only be used when of course the base of the logarithm is greater than Z and not one but also when the arguments of the logarithms are greater than zero so let me show you two examples of how we can use this product rule of logarithms let's say we had log base 3 of 7 plus log base 3 of five because we have two logs being added and each of those logarithms have the same base those can be Rewritten as a single logarithm by keeping that same base of three and multiply their arguments 7 * 5 and 7 * 5 is 35 and let's do one more this time let me make it a bit more difficult I've got 2 Log 5 plus a half log 16 now in the product rule notice the coefficients of the logarithms are all one and in this example they're not one so we have to take care of that by doing the power rule of logarithms which tells me I can take each of those coefficients and write them as the exponents on the arguments so this would be equal to log of 5 sared which is 25 plus log of 16 to the half 16 to the half just means otk 16 which is four and now I can combine these as a single logarithm because they're logs being added to have the same base right the base of both of these common logarithms is 10 so I can combine them as a single log by keeping that base of 10 and multiplying the arguments 25 * 4 and 25 * 4 is 100 and this would be easy to actually fully evaluate log 100 means what exponent goes on 10 to get 100 that's two the second rule for this section is the quotient rule of logarithms the quotient rule states that if you have two logarithms being subtracted so log base B of M minus log base B of n as long as the bases of those logarithms are the same we could rewrite those two as a single logarithm with that same Base by dividing the arguments so m / n and once again this rule can only be used as long as the base of the logarithm is bigger than Z not equal to 1 and the arguments of both of the logs are bigger than zero and now let me give you a quick example of how this can be used let's say you had log base 4 of 30 minus log base 4 of 6 two logs being subtracted where the logs have the same base you can use the quotient rule which says keep the base the same and divide their arguments 30 / 6 that's 5 and this expression means what exponent goes on four to equal 5 so it's 1 point something you use your calculator to get an approximate value it's about 1.16 must know number six other rules and tricks let me show you a few other rules that are going to be helpful when working with logarithms and before I show you the rules let me remind you about the restrictions on a logarithmic expression the base of the logarithm has to be bigger than zero and not one and the argument of the logarithm has to be bigger than zero the first rule I want to show you is is the change base formula which allows you to change the base of any logarithm so if we have log base B of M and we want to change the base from B to some other number we can do that by writing it as a quotient of two logarithms with that new base that we want so I could rewrite log Bas B of M as log base a of M over log base a of B where this a could be any number bigger than zero and not one so let me show you how this formula Works let's say you had log base 4 of 20 you could rewrite that as a quotient of two logarithms with a new base and most often this formula is used to change the base to 10 so it becomes the common logarithm so this would be log base 10 of 20 / log base 10 of 4 so log base 4 of 20 can be Rewritten as log 20 / log 4 that's the change base formula another rule that might be useful is the rule that tells you that b to the power of log base B of M that would just equal m if you think about what log base B of M means that means the exponent that goes on B to get M so if you take that answer and put it as the exponent on B you of course get M so an example of how this would work is that 3 to the power of log base 3 of 9 well that would just equal 9 this exponent log base 3 of 9 means what exponent goes on three to get 9 and that's of course two so when you put that as the exponent on three 3^ s is 9 so as long as these two numbers match then the argument is going to be the answer to this expression and this tool is often used to rewrite a number as a power with a different base so for example if I wanted to rewrite five as a base 2 power well I could change this to a power of two if I make the exponent log base 2 of five and then a couple small other tips if you have log base a of a where the base and the argument match each other well that's just going to be equal to one because a to the^ of 1 is equal to a so an example of this would be if you had log base 5 of five that means what exponent goes on five to get five that's one and another small rule log base B of one well anytime the argument of your log is one we know the answer to the log is zero for example if you had log base 5 of 1 what does that mean it means what exponent goes on five to get one well the answer to that is 0 because 5 to the^ of 0 is 1 must know number seven solving exponential equations let me write three exponential equations and I'll show you the different strategies you can use to solve them if we look at the first equation on the left notice all we're trying to figure out is what exponent goes on two to equal 20 well we can just convert this to a logarithmic equation by saying the exponent equals log base 2 of 20 this function will find the exponent that goes on two to get 20 it's going to be 4 something to find an approximate value you can either type it like that but if your calculator is not able to change the base you can use the change a base formula to convert this as a quotient of common logarithms so we could write that as log 20 / log 2 and it's about 4.32 the next equation notice we have two terms that have a variable in the exponent when that's the case when you have a power equals a power try and rewrite the power so they have the same base so I would rewrite that 8 as a base 2 power I can change the 8 to a 2 cubed and then using exponent rules I know I can multiply that 3 by that 2x - 5 3 * 2x - 5 would be 6x - 15 and once you have power equal power where the bases are the same the only way this equation could be true is if the exponents were equal to each other so you can just set the exponents equal X = 6x - 15 and then solve this by rearranging and then the last exponential equation looks very similar to the middle one where we have a power equal to a power but it would not be easy to rewrite them with the same base so what we do instead is we take the common logarithm of both sides and then use the power rule of logarithms which tells you you can take the exponents of the arguments and write them as the coefficients of the logarithms on the right side of the equation we can do 2x - 5 * log 7 by multiplying log 7 by both the 2X and the -5 and then I'll rearrange this equation to get the terms that have an x to the same side of the equation a couple things to do here I could simplify this expression by taking this coefficient of five and writing it as the exponent on the argument 7 to ^ of 5 is 16807 and on the right side of this equation I notice both terms have an X so I'll common factor out an X and then this factor I can simplify I'll move the two as the exponent on the seven to make it 7^ s which is 49 and then when subtracting Logs with the same base you can write it as a single log by dividing the arguments so that would become log 49 / 2 I'll make a bit more room and my last step to isolate the x is to divide both sides by log of 49 /2 and then using the change of Base formula I could rewrite that as a single logarithm I could write it as log base 49 over2 of 16807 that's the exact value of this answer using a calculator you could get an approximate value of about 3.04 must know number eight solving logarithmic equations let me write three log equations to solve let's look at the first equation one General strategy when solving a log equation is to get the equation to be in the format log equals log because if we have two logs equal to each other where the bases are the same then you know that the arguments must be equal to each other so this equation looks like it's already almost in that format if we take this coefficient of two and write it as the exponent on the argument then that would make this equation easier to solve so I'll write this as log of x + 5 equal log of x -1 2 and now that we have two log Expressions equal to each other where the bases of those logs are the same the bases are both 10 I know the only way this equation would be true is if the arguments were equal to each other so I can say that x + 5 would have to equal x - 1^ 2 and now this is just a quadratic equation to solve let me expand the right side x - 1^ 2 we do x -1 * another X - 1 that would be x^2 - 2x + 1 rearrange this to set it equal to zero and I can Factor this quadratic by finding numbers that multiply to -4 and add to -3 the numbers that satisfy that product and sum are x - 4 and x + 1 using the zero product rule I know the product of these two factors would be zero if either of the factors would be zero this Factor would be zero if x was 4 and this Factor would be zero if x was1 so my solutions to this equation are x = 4 and X = 1 now we have to be careful when solving logarithmic equations sometimes you might generate what's called an extraneous solution which means a solution that might work for many of the equations throughout our process but does not work for the original equation and when we use log rules remember the Restriction we have for logarithms is that the argument of a logarithm has to be bigger than zero so we need to check both of our Solutions and make sure they don't make any of our logarithmic Expressions become undefined make sure it makes the arguments of both our log Expressions be greater than zero four subbed in for both of the x's in my log Expressions well it would make that argument N9 and this argument three those are both greater than zero so that's fine but if I sub in -1 it makes this argument four but it makes this argument NE -2 so this answer of -1 makes this log expression undefined which means it cannot be an answer to this equation so we actually have to reject that answer and say that's an extraneous solution the only answer to this equation is four now the fact that this makes both of these arguments positive doesn't guarantee it's the right answer you should also take the time and make sure that when you sub it in for the X's it makes the whole left side equal to the whole right side this middle equation is in a format that if we convert this to exponential form it's very easy to solve so if you have a logarithm equals a constant just convert to exponential form this log expression means that 2 is the exponent that goes on five to equal 2x - 3 so I can write that and now just solve this linear equation add the three to the left side and then divide the 2 and you get x = 14 checking this answer in the original log expression it makes the argument here be positive and if we actually evaluate the whole left side by subbing 14 for X we get log base 5 of 25 which is 2 so 14 is the right answer and the last log equation I see two terms that have an X so let's get them on the same side of the equation so I'll take this negative log add it to the other side and now two logs being added where the logs have the same base I can use the product rule to combine those as a single logarithm by multiplying the arguments multiplying the arguments gives me x^2 + 9x + 18 now that it's a log equals a constant let's convert it to exponential form remember on the left we have the common logarithm so the base is 10 so that equation means that 10 ^ 1 is equal to the argument it's a quadratic so I'll set it to zero before solving further and then that's factorable the numbers that multiply to 8 and add to 9 are 8 and 1 so the solutions to this equation are x = 8 and x = -1 this answer of X being -1 makes that argument POS 5 makes that argument pos2 so because both those arguments are bigger than zero that is probably a fine answer you should also double check and make sure it makes the whole left side equal to the whole right side but if you sub 8 in for the X in the original equation it actually makes both of these arguments be negative so it makes both of those expressions undefined so it cannot be an answer to this equation so xal 8 is an extraneous solution so the only answer to this equation is x = -1 must know number nine what are logarithms used for there are two main things I want to show you the first of which is that logarithms can be used to help analyze very small or very large numbers for example let me give you a set of values I'll start with very small number numbers like 0.1 and then I'll increase by factors of 10 until I get up to some very large numbers so let me just continue this table so you can see at the extremes of this table I've got some really small numbers at that end and some really big numbers at this end so if I were to take each of these numbers and do the log of each of these numbers so I'll take all of those numbers and put them as the input into this log and then let's look at what is this logarithm output if I take this first number and do the log of that number that means what exponent goes on 10 to get this number 10^ the4 equals that number so that number would translate to -4 the next number input it as the argument of the log and I get -3 and then as I continue to sub these numbers in notice when I take this set of very small and very large numbers do the log of each of them I get this set of numbers which would be a lot easier to work with and there are a few formulas that take advantage of the fact that a logarithm can output these nice manageable numbers for example the pH formula is equal to log of the hydronium ion concentration how is this formula used well pure water has a hydronium ion concentration of 1 * 10^ -7 so really small number 0.000000001 what we do is we calculate the pH of Pure Water by doing the negative log of its hydronium ion concentration 10 to the7 and then evaluating this is just what exponent goes on 10 to get 10 -7 well that's -7 so we Doga -7 which is just 7 pure water is neutral so this seven is actually unitless it's just a way of taking a hydronium ion concentration which does have units in moles per liter but converting that really small number into a number that is easier to analyze and there are lots of other formulas which take advantage of the fact that logarithms can do that with really small or really big numbers for example there's a formula for sound levels measuring sound in decb their difference in their sound levels is equal to 10 * log of the ratio of their sound intensities let me make a little bit more room and then talk to you about a logarithmic scale on a graph sometimes logarithmic scales are used when graphing data with a very large spread when graphing data or functions usually use a linear scale but there's another scale you can use called a logarithmic scale sometimes a logarithmic scale is used when graphing data with a very large spread of values for example let's look at the function y equal 10 ^ x if I were to make a table of values for that function I'll choose X values from 3 to 3 and if I calculate the Y values by doing 10 to each of those exponents I get the following y values now let me try and graph these points to show what this function looks like your initial Instinct would probably be to use a linear scale so let me make a little bit of room and we'll do that if we zoom in on this graph notice that the scale I made on the y axis the numbers are going up by a common difference each time or another way we should say that is that measurements that are an equal distance apart like 10 and 20 and 50 and 60 those are an equal distance apart their difference in values is equal they both have a difference in values of 10 and now if I were to take all of these points from the table and try and get them onto this graph these first few points are going to be very difficult to plot because they're so small and this point at the end is so big it's not even going to fit on the graph but I'll do my best to plot them and if I connect them to show it this function looks like I get this exponential curve and there actually is a horizontal ASM toote right along the xaxis at the line Y equals 0 so I'll draw that in as well but me showing you this linear scale for this function uh is just to point out that seeing the difference in these values is very difficult to see and because these differences are so big we're going to stop being able to get points that fit on our graph so there's actually a different type of scale we could use for a function just like like this and it's called a logarithmic scale so let me make some room and do another graph now let me zoom in and label some points on here so this logarithmic scale I made there's a few things I want to point out first of all pairs of points that are an equal distance apart like 1 and 10 or 100 and a th they're an equal distance apart but their difference is not equal this difference is 9 but this difference is 900 the difference isn't equal in a logarithmic scale but the r ratio of the numbers is equal for points that are equal distant apart so those points that are equal distance apart the ratio of 10 ID 1 is 10 and the ratio of 1,00 divid 100 is 10 so in a logarithmic scale these numbers are going up by a factor of 10 each time multiply by 10 you get to the next number and it doesn't have to be by 10 but that's a common one that is used another thing to know that if you were to do the common logarithm of each of these numbers the values that would get would have an equal difference right logarithm of this would be -3 then -2 then1 then zero then 1 2 3 4 the other thing you need to notice is I didn't draw on an x- axis because that would imply we were at a yv value of zero and if you think about it what would be the next y value as we go down well we would just be adding another zero in the decimal place before the one so it would be 0.00001 and then the one after that would have four zeros than the one and so on so the numbers will just keep getting closer and closer to zero but they will never get there so for that reason we can't draw in an x-axis and when I plot each of these points on this logarithmic scale let's watch what they look like they actually form a perfectly straight line and I was able to plot every single point that is in that table on my graph here must know number 10 the derivative of a logarithmic function let me start by showing you the derivative rule for the derivative of log base B of X the derivative rule tells me this would be equal to 1 / X Ln of B so you take the argument of the log it goes here and you take the base of the log and that becomes the argument of the natural logarithm and now let me show you how this would work if we had a composite function so if we wanted the derivative of log base B of FX so if the argument of the logarithm is a function of X you need the chain rule which would tell you that you would do 1 / the argument which is FX time the natural logarithm of the base but that has to be multiplied by the derivative of the inside function so it have to be multiplied by frime of X so an example where we would need this is if you wanted to find the derivative of two * log of 3x^2 so the first thing I could use the constant multiple rule I have a constant factor I could just move that in front of the derivative and because the argument is a function of X I need the chain rule so this would be equal to 2 multiplied 1 over the argument which is 3x^2 * the natural logarithm of the base of the log and I don't see a base so it's the common logarithm so the base is 10 and that has to be multiplied by the derivative of the argument the derivative of 3x^2 is 6X and then this could be simplified I've got a 6 / 3 that's two and this x can cancel with one of those X's so the final answer in the numerator there's 2 * 1 * 2 that's 4 and in the denominator is just x * l 10 so there you go hopefully that video has made you more confident in your understanding of logarithms let me know in the comments what you want me to do a video about