Indices & Logarithms Full Lesson | Laws, Change of Base & Exam Questions Explained

Añadido: 2026-06-02

[00:00:00]Hello everybody, welcome to this YouTube channel. My name is Jacob Sandra. So did you know that a strong understanding of indices and logarithms can improve or can transform your understanding in mathematics. In this video, we're going to look at logarithms and indices. So first thing first, when it comes to logarithms, there are few properties that you need. You can see we've got three properties that I've put here which we need in order for us to work out exam questions for this one. Indices they are also properties and please keep this in your mind. These are not the only properties we need in order for us to answer exam questions. One quick thing change of base. Under logarithms there is what we call change of base property. So when you've got different bases how do you get to solve that um question for this one the bases are the same. So the way of solving it will differ from how we solve this one. That's why it's called change of base for this one. Let's quickly talk about the properties of logarithms and after that we answer these questions including this one. So the first property here says that when you are multiplying okay this is multiplication x * y the multiplication of logarithms changes to addition. So you can see log x + log y because we are multiplying.

[00:01:47]Let me just write in an example. Let's say you've got log of a b. Log of a * b. This one of course will give us a log of a plus log of b.

[00:02:09]So um multiplication changes to addition.

[00:02:16]For the next one here, this says that when you are dividing two elements or we say the division of logarithms, log of x / y changes to subtraction. So this is log x minus log y. Remember these properties. So this is number one. Number two, for example, if you've got log of a over b, this of course will change to log of a minus log of b. So this is how it changes.

[00:02:56]And then for this one, it says that when there is an exponential involved, when we talk about exponential, we're talking about the power. So there is base and the power the moment you introduce log this exponent will come in front like it is this is how it will look like. So what it means here it means when you've got for example a power b the moment you just introduce log this will be b log of a so this will come in front like it is here that's how uh this state property works let's now look at the question or the question so to say and uh see that we work them out. So the question that I've just erased here, let me write it here.

[00:03:53]It's uh log base 2 is it two? Yeah of x + log base 5 of x = 4. This is an equation we we need to to solve. Okay, I hope you'll be able to to follow through all the way up to the end.

[00:04:16]So let's clear up here so that we can get to see how these logarithms are worked out.

[00:04:28]So the first question here I want you to see that we've been given 7^ x + 1 and they're saying this is equal to 13 - 4.

[00:04:45]So here we can use our simple mathematics. What we know we have 7^ x + 1 is equal to a 9.

[00:04:56]13 - 4 will give us a 9. Now this is where the importance of logarithms comes in.

[00:05:06]We can't balance the bases. There is what we call balancing the bases using this property here. But here seven is not uh cannot be written in uh something such that the same come comes this side we balance the same thing. I don't know if you understand what I'm trying to say. So in other ways I'm just trying to say we cannot balance the bases. So when you've got an equation that looks like this where you can't balance the base this is where you introduce the log. So log 7^ x -1 is equal to log 9. So you introduce logarithms to both sides of the equation. This side and this side. And this is how it will look.

[00:05:56]And when you introduce that, you can see this property which says that when you've got an exponent, it should come in front like this. And if you can see, we've got five and this exponent. So this will be x + 1 and then log over 7 is equal to log log of 9. So the property that we have here has been applied right over here because the exponent has come in front.

[00:06:31]Now here we do our normal mathematics where we we distribute. So this will be x log 7 + log 7 is equal to log 9. So let me explain what I've just done here. So what has what has happened here is that there are these brackets. So we've multiplied this log 7 by x. It's x log 7. log 7 * 1 it's log 7 which is equal to log 9 which is here. So from here we can group the like terms. We've got x log 7 is equal to log 9 minus let me clear up. Let me do this so that we get to our space.

[00:07:29]Let me clear up here so that we can get to our space and follow through what we are doing.

[00:07:38]So if you're checking nicely, you will see that here we've got brackets.

[00:07:48]Here we've got open bracket x + 1 and then here log 7 is equal to log 9. I've just copied this.

[00:08:04]So from this point we can of course distribute this will give us x log 7 which is uh plus here log 7 which is equal to log 9. So you just distribute this log 7 * x it's x log 7 log 7 * 1 it's log 7.

[00:08:35]So we group the like terms. So every term with x will remain which is equal to log 9 minus log 7. So if you really understand the properties of logarithms, there is a property that must be applied here which is this one. Can you see there's minus minus sign here. So meaning it will be this divide by this.

[00:09:04]And by the way, let me just say something which is very important here.

[00:09:09]When it comes to logarithms, you are you something that looks like this. Log a over log b. This one is not the division of logarithms. It's not like this or what we divide. We divide these variables inside or numbers. Note the logide by log. The moment you see that there is logide by log. This one you can write it as log base b of a.

[00:09:38]This one. Okay. It's also a property.

[00:09:42]Log whatever is here becomes the base.

[00:09:46]This one becomes what you're finding for log base b of a. So now here I can apply this property. It will be x log 7 is equal to log 9 / 7. I hope you're able to see now that this property has been applied right over here. Now here we can divide by log 7 log 7. So what we have now is x is equal to because these two cancel log of 9 / 7 over log of 7. And if you're able to see nicely there is a log here and a log here. So this property can be applied.

[00:10:40]We can therefore say x is equal to log base 7. This is 7 because this b became a base. 7 is here. And where there is a here we are going to write 9 / 7. So this is our answer. Log base 7 of 9 / over 7. That's a solution for this.

[00:11:08]And I've explained why we applying log here. Why didn't we balance the bases?

[00:11:15]We cannot balance the bases because the bases are not balancable. So in that situation the only thing that can help us it's the logarithms because there is one property which is this one which says that when you've got this exponent it can come in front like this x has come in front just like that. So let's look at the next question and see how that question which now has got already the logs how it's supposed to be solved.

[00:11:47]So when you've got a question that has got a log or log logarithms involved already, you are supposed to to follow also the steps of logarithms.

[00:12:00]Okay, whichever equation that you have under logarithms, you just need to know how to apply the property. Which property you're supposed to put here, which property should be put there. So we've got log base 2 over x + 2 which is equal to 3 + log base 2 of x. So with equations just the way normal equations are done the first thing that you do with equation equations it's either you get rid of the brackets or you group the like terms if there are no brackets involved. But now with logarithms here, if you're able to see there are these brackets, this does not mean you need to multiply this log by everything that is inside. No, doesn't work like that. Okay? Whatever is in front here of the log like this should be left just like that. So what we're going to do here will simply be to group the like terms. So every term with log will be on one side of the equation.

[00:13:15]So log base 2 of x has come this side it's plus it has now turned into a minus and we've remained with the three. So just by observing you can see that there is a property that can be applied right over here this property the division property. So when we apply it, it's going to be log base 2 of x + 2 / x. So this is the division log. So what is here divide by what is here. This is how it looks. Even here what is here divide by what is here. And this whole thing will give us a three.

[00:14:01]So all what we are moving towards here it is to get rid of the log equations or should I say equations that involve logarithms you can't solve minus getting rid of the log. So this log base 2 must be gotten rid of. And how do we do this?

[00:14:22]So now look at this. when you've got log base a okay of b is equal to c. So now like it is here a is here and then b is here which is here is equal to c where three is. So this one can be written as b is equal to a power c. Have you seen this one is the same as the as this?

[00:14:50]Applying also this property it will take us back to this. So now what will happen here? This will just be x + 2 over 2 is = 2^ 3. This will come and push this. It will become a power like it is here. This a pushed c. It was a base here and c is an exponent. So here we can say x + 2 / 2 is = a 9. We group the like terms. I mean 8. Sorry, this is 8 is equal to 8. So I've got x + 2 is = actually here it's not two, it's x.

[00:15:39]x this is x. The x which is here h 8 x we solve this. It will be 2 = 8 x - x.

[00:15:49]So from this point, let me bring it here. From here, we can say we've got 2.

[00:15:56]We've got 2 is = 7 x / 7 by 7. This and this will go x will simply be equal to 2 / 7. That's the value of x for this one.

[00:16:12]So just follow the steps. Step number one, group the elements and this is how it will look. For number two, check the property that you can apply. I've applied this one and this is where it reached me. From this point, I I applied this idea where this base will come this side and whatever is here becomes an export. And this is what happened. So this came here and then this whole thing it was here x + 2 / x and then we crossed so we did some cross multiplication here because of the fraction there is an invisible one here so x * 8 it gave us 8 x and then 2 um x + 2 * x it gave us the same thing. We group the like terms and this gave us the solution that we are looking for of x to be equal to 2 over 7.

[00:17:11]Let's go to the next question. We check the next question here.

[00:17:15]For the next question, this is now where we need to apply change of base.

[00:17:24]So stick around. Don't go away. You need to check and see and also understand because this will help you be able to understand some more concepts in mathematics because laws of indices and logarithms they are very applicable in any advanced mathematics that you're doing.

[00:17:51]So here this is how the question that we want to do now looks. We've got uh log log base 2 of x plus log okay base 5 of x which is equal to four.

[00:18:10]So basically here we cannot do anything because the bases are different change of base we have to balance the bases. There are only two bases or two types of bases. There is base two and base five. So we can decide which base we want to take this equation into. If you want, you can take it to base two or if you want, you can take it to base five. So in my case, I'm going to take it to base 5. So taking this to base five, we need a property. Let me put it here. We need a property which says that log okay base a and then I mean log base b of a if I want to change this to take it to base c because it's in base b I'm going to write it as a log base c where I'm taking it of a what is here and then say over log base c where I'm taking it over the base for previous base. I hope you're able to see that this is a property that we need to apply. So since this we want to take it to base 5, we're going to say log base 5 of x over log base 5 of 2. Log base 5 of x over log base 5 over what is here? Two. and then say plus log base 5 of x which is equal to a 4.

[00:19:53]So we've done that. That's the first step. Now after doing this you have to break this fraction. And how do you break the fraction? You break this fraction by simply multiplying everything that is inside by log base 5 of 2. So when we multiply log base 5 of 2 * log base 5 of x over log base 5 of 2 this will cancel we remain with the log base 5 of x.

[00:20:26]Then here log base 5 of 2 * log base 5 of x. This will just give us log base 5 of x * log base 5 / 2 which is equal to here when you multiply it will be 4 log base 5 of of no no yeah log base 5 of 2. So it will just be this time this it will just be 4 log base 5 of 2. And from this point of course, let me clear up.

[00:21:04]From this point where we are, I want you to see very carefully.

[00:21:11]We can use the idea of factorization.

[00:21:13]Okay, I know some of you may be thinking but we can see plus here. Why can't we apply this property? It cannot wait. we can't apply because here let me copy paste what we have so that you see clearly what I'm talking about.

[00:21:32]So here we've got log base 5 of x + log base 5 of x times. So in case you don't know multiplication we sometimes use a dot to mean times log base 5 of 2 is equal to 4 log base 5 of 2. I've just copied this.

[00:22:01]Now from uh where we are here what we can do here we need to factoriize to factor out what is common because this property addition cannot be applied here. We've got multiplication here. So this one if it was not there we can use that property but now it's here. So the best thing that we can do here is to factor out we factoriize what is common. So log base 5 of x into log base 5 of x it will just give us a 1. log base 5 of x into log base 5 of x * log base 5 of 2. It will just be plus log base 5 / 2 and close. So now from here we can say equal to log 4 log base 5 of 2. We now divide okay by 1 + log okay base 5 of 2 both sides 1 + log base 5 of 2. So these two go we will remain with log base 5 of x which is equal to uh 4 log base 5 of 2 over 1 + log base 5 of 2. So if we want here we can apply this property. Have you seen the property which is here? Here we can even write it like this. Log base 5 of x is equal to log base 5 of 2^ 4. So this one goes there as an exponent applying this property.

[00:24:00]So it will be over 1 + log base 5 this one. So now let me clear up so that we can see what we have.

[00:24:17]clear up here.

[00:24:19]I hope you're able to see and follow what is happening.

[00:24:32]So now from here we can say log base 5 log base 5 of x is equal to log base 5 / 16. So here 2^ 4 will give us 60 and then you say over 1 + log log base 5 log base 5 of 2.

[00:25:05]So that's it. Now here if you can check nicely there is this property here which we applied. You can see that now we want x to be alone. We want x to be independent be alone. So we're going to apply that property.

[00:25:24]Whatever base is here will come and push this all of it. It will become an exponent like it was here. C is an exponent because a came here. So five will come here and all of this will become an exponent. So this will be x is equal to 5. Okay. And then we say power log base 5 of 16 over 1 + log base 5 of two. So this is a the value of x for this question where we applied change of base. So now let me just run you through a bit with change of base. We check at the the bases. These are called bases.

[00:26:13]If they are different, you are going to pick one of the terms. If you want, you can pick this one or this one. You change its bases or you change the base.

[00:26:26]In my case, I decided to take it to base five. So that base five, it was here.

[00:26:31]This one I took it to base five. We used this idea right here. This is how we change to BA base something. And this is an idea. So we took it to BC like it is here and we came up with an equation which we solved. We worked out and also here we applied now this idea. I hope this uh makes you understand.

[00:26:55]Now let's clean up. We go to the laws of indices. Okay. So for the laws of indices laws of indices there are few things that you need. There are very few things that you need in order for you to find mathematics easier to work with.

[00:27:18]Okay.

[00:27:20]So the few things that you need or some of the few things that you need are these common properties.

[00:27:27]So the properties that we have here the first one says when you are multiplying same bases remember with indices what is down here these are called bases and what is on top here they are called exponents okay or index. So this is an index this is also an index. Now when you are multiplying the same base like it is times the same base don't mind very much just get that base and then add the powers or the exponent n + m. This is how it looks. So this property is called multiplication property under indices.

[00:28:11]For this other one, it says when you are equating the same bases, regardless of how the exponents are or the powers are, uh, for as long as the bases are the same, they are equal to each other. This simply means the exponents are also equal to each other. No matter how they they look, even if they look different, since they are coming from the same bases, equal bases, and those bases are equated to each other, so you can get the exponents and say they also equal to each other.

[00:28:47]The other property is division.

[00:28:51]When you're dividing the same bases, you subtract the top power uh and you subtract top power minus the bottom power or bottom exponent. And also this property we call it power to power. When you've got an exponent here and another exponent here, even another another you multiply them. So this time this this is how it will look. Let's uh take our time and answer these three questions. So the first question here they've given us a 2 x + 3 is = 1 / 16.

[00:29:31]So here there are properties that you need to to consider. If you want you can say okay I can apply this property because this one looks like this. So I can drive it back this one. But I'm not going to show you that. I'll show you something that is very simple. So when there is an exponent here which is a one don't mind very much. Balance the bases. So this will be 1 / 2^ 4. I know you know that when you say 2^ 4 it will give you a 16.

[00:30:09]So from this point we can now say 2 x + 3 is equal to we apply the property that looks like this. when you've got a 1 / a power n this will just be a one I mean it will be a power n so this fraction I mean negative exponent property is what is applied here so whatever you have here as a denominator it will make a move go on top or you sprocket one comes down this one goes was on top.

[00:30:50]Meaning after just doing that movement the exponent will change each side. It was positive here it will be negative and this will just give us a power n. So here when you look at this it will just be a 2 power -4.

[00:31:11]I hope you're able to see that. Now from this stage we can now apply this property because the bases are equal and if the bases are equal you equate to the exponents. So this base and this are equal meaning we shall have x + 3 = -4.

[00:31:32]So this is x = -4 - 3. x will be equal to -7.

[00:31:40]This is our answer for the first question.

[00:31:44]We go to the next question here. So for the next question, we do the same. We have to balance the bases. That's the first thing for this one.

[00:31:55]So here it will of course be we balance the bases that we we have right here.

[00:32:10]And what it means when we say balancing the bases, it means we make sure that these bases are equal to each other.

[00:32:19]Here I've got 27^ x which is equal to 81.

[00:32:25]Now we need to balance the bases. So how are we going to balance these bases? We think of an a number. In my case I've thought of a three and then power three.

[00:32:36]I know that when we say 3^ 3 that is a 3 * 3 * 3 will give me 27. So meaning even this side I should write a three so that we can balance the bases like it is here. So it should be three this one power something that will give me 81 and that is a four because 3^ 4 will definitely give me 81.

[00:33:02]Now we've got a problem. The problem is right over here.

[00:33:07]Have you seen these brackets? We need to get rid of the brackets. Now, how do we remove these brackets? We follow the property this property which says that when there are brackets like this power to power, you multiply them. So, this will be a 3 power 3 * x is equal to 3^ 4. So now from this point we can now use this property. This property is commonly or is one of the commonly used property.

[00:33:40]So the bases are equal to each other. I can say 3x is equal to a 4ide by 3 by 3. This and this will go x will simply be = 4 / 3. That's our solution for this one.

[00:33:59]We go to the last question which is five and then we've got uh the exponent there 1 - 3 x.

[00:34:15]So for this one right here we we apply the laws of indices. So the first thing the way it looks we have to balance the bases. Okay? So we need to balance up the bases. So the bases here for them to be balanced since we've got 5^ 1 - 3x is equal to 125. So for us to balance the bases here, we're going to write this as 5^ 1 - 3x is equal to 5^ 3 because 5^ 3 will give us a 125. So the bases are equal to each other. You can see that we're jumping on this property.

[00:35:00]We can say 1 - 3x = 3. 1 - 3x = 3. So here it will be -3x = 3 - 1 of which when we work out here it will give us a -3x is = a 2ide by -3 by -3 these two go x will be equal to -2 over 3. So that's uh that is that's the answer. Thank you so much. I hope you've picked one or two things out of this video. You might be that person who is interested to get in touch with us. So this is the number you can WhatsApp or call. We will definitely be able to advise on how you can be part of the team JS Lenny Academy online teachers. Thank you and bye-bye.