 Hi, I'm Zor. Welcome to User Education. Today, we will solve some problems about logarithms, which have been explained in the previous couple of lectures. We're talking about definition of logarithms, main properties, graphs, etc. So, this lecture and maybe one or two others, we will develop to solving problems with logarithms, which is extremely important to basically understand the concept much better. This lecture will be about very simple problems, which are very, very basic. They are based on the definition of the logarithms. So, allowing me to basically spend some time on this because I believe it's very important to really feel how the whole thing with logarithms is working before we go through anything more complex. So, simple problems. I have a set of problems which are formulated as follows. Given the following equality, what's necessary is to rewrite this equality in exponential form. So, basically, we have to think about what logarithm actually means. What's the definition of the logarithm? Okay, the logarithm of 9 with a base 3 is the power, we have to raise 3, the base, to get 9. Now, obviously, this is 2 because 3 to the power of 2 is equal to 9. So, basically, these two forms are equivalent to one another. And to convert one into another, you really have to understand what logarithm is about. So, again, since this is a power which 3 is supposed to be raised to get 9, and obviously 3 to the second power 2 is 9, so that's why it's equal to 2. Similarly, logarithm of 8 base 1, 2, one second, now, this is minus 3. Why? Again, let's think about the definition of the logarithm. It's supposed to be the power, I have to raise the base which is 1 1 half to get 8. Okay. Is it true to the power of minus 3? Will it give me 8? Well, obviously it is true because minus means we have to reverse the 1, 1 half into 2 and 2 to the power of 3 would be 8. So, this is an equivalent to this. Let's go further. Log base 1 tenths of 1 thousandths is equal to 3. Again, if I will raise the base, which is 1 tenths, to the power of 3, I will get 1 tenth times 1 tenth times 1 tenths of 1 thousandths. So, this is justification for this. That's why this can be rewritten as exponential form. These two are completely equivalent. Next. Okay, let me wipe out this. So, we'll preserve the large format. I have three more of the same type. These are illustrative problems just to show what's the definition of the logarithm actually is. Log base 1, no, sorry, base 2, 1, 10, 24 equals minus 10. Well, what does it mean? It means 2 to the power of minus 10 should be equal to 1024, right? 1, no, sorry, again, it's 2, not 1 half. To the power minus 10 equals 2. Now, what is 2 to the power minus 10? Minus means 2 should be inverted into 1 half and then 10 times multiplied by itself. Well, you can verify it is 1024. Well, I was working with computers for a very long time and in computers powers of 2 are very important for some reason. So, I just remember that powers of 2 I remember up to probably 12 or even more. So, these are completely equivalent. This means that 2 to the power minus 10 is equal to 1024 and this means exactly the same thing. Paulette, a little bit more strange-looking log-based square root of 3 of 1 third. What is this? Minus 2. Well, let's think about it. If I will have minus 3 to the power, square root of 3, sorry, to the power of minus 2, what does it mean? Well, minus means I have to invert it. It would be 1 over square root of 3 and then square, well, square root of 3 square would be 3, so it would be 1 third, which is exactly this. So, again, this means that square root of 3 to the power of minus 2 is equal to 1 third and this means exactly the same thing. And the last one in this series, log 3 of 1 over square root of 3 is equal to minus 1 half. Well, let's do this. 3 to the power minus 1 half. What is it? Well, 3 to the power of minus 1, let's say it would be 1 third and then 1 half means square root. So, it would be square root of 1 third, which is 1 over square root of 3. So, these two forms, these two forms, these two are completely equivalent to each other. They are actually a definition of the logarithm. I didn't calculate anything. I have just rewritten one into a different format. Meaning of these two things is exactly the same. This means that the 2 to the power of minus 10 is equal to 10 to the power of 4. This means exactly the same thing. 2 to the power of minus 10 equals to 1 over 10 to the power of 4. The same meaning, different notation, no more than that. What I actually wanted to emphasize is that logarithms are just a different notation for exponential function. Nothing more than that. All right. So, this series of problems we have finished. Next one is actually very similar, but in reverse. Now we have to rewrite in the logarithmic form something which is given as exponential. So, if I know that 10 cubed is equal to 1000, how can I rewrite this exponential expression in logarithmic form? Okay, let's just think about it. I know that 10 to the power of 3 is equal to 1000. Right? Now, that means that if I will get logarithm base 10 of 1000, I should get 3. That's what it means. This is a definition of the exponential expression, basically, notation of the exponential expression. This is exactly the same expression written differently, written using the logarithms. The meaning of this notation means exactly the same as the meaning of this notation. Again, 10 to the power of 1 half is equal to square root of 10. Or, alternatively, we can say that log base 10 of square root of 10 is equal to 1 half, which means 10 to the power of 1 half is equal to square root of 10. To the third degree is equal to 8 means same thing as log base 2 of 8 is 3. So, the base here becomes the base here. The result of this raising to the power is supposed to be under the logarithms and the exponential part itself becomes the value of this logarithm. 8 to the power of 1 third is equal to 2, which means log base 8 of 2 is equal to 1 third. 1 third is actually a cubic root. So, cubic root of 8 is 2, obviously. 1 half to the power of minus 3 is equal to 8. It's exactly the same as log base 1 half of 8 is equal to minus 3. Both are correct indeed. Minus means we have to invert 1 half into 2, basically, and then cube would be 8. Same thing there. And the last one in this series, 4 to the power minus 1 half equals 2 1. Well, is it correct, by the way? 4 minus means it's 1 fourth. 1 half means square root. Square root of 1 fourth is 1 half. Exactly. So, in the logarithmic notation, it means log base 4 of this equals whatever the exponentials in this. Okay, that's the last one in this series. So, we have converted logarithmic expression into exponential and exponential into logarithmic. In both cases, my most important point is that these are exactly equivalent to each other. Just different notation of the same thing. Now, let's assume that all logarithms below are base 1 half. So, what's the logarithm of 1 half? Here's an interesting consideration. When you have an exponential function, let's say 2 to the fourth is equal to 16. You can just calculate. If you don't know this value, you can calculate it. Well, you multiply it 2 by 2 by 2 by 2, you will get 16. This is a calculable thing. What if I will say an equivalent thing? What is log 16 2? Well, here you have to make a little guess, basically. Well, which power should I use to raise the 2 to get 16? Well, you just think about, well, 2 power of 2 would be too small because 2 square would be only 4. 3 also too small. 2 to the third power is 8. Finally, 2 to the fourth power, yes, this is 16. So, you are kind of guessing the answer to this problem. You cannot directly calculate it using whatever your... I'm not talking about calculators. I'm talking about just using your mind. So, you really have to kind of guess that this is 4. You have to try and fail and try and fail and finally get the result. Now, this is a very important difference between logarithms and exponential functions. Exponential functions in simple cases. I'm not talking about exponential functions of a complicated case like this. No, I'm talking about integers, basically, which are relatively easy to calculate. Integers and inverse two integers things. Now, this is something which you really have to, well, think a little bit. Just believe me. Now, which power should I use 1 half to raise 1 half to get the 1 half? Well, obviously, in this case, it's an easy answer. It's 1. Indeed, 1 half, let's just make, you know, it's checking, basically. To the first power means it's just by itself, so it's 1 half. So, this is correct thing. Since this is correct, this is equivalent to this, then this is correct. Next, power, which power should I use for 1 half to get 1? Well, again, this is an easy thing because I know that any number raised to the power of 0, by definition, actually, the power of 0 is 1, right? So, I know that 1 half to the power of 0 is equal to 1. So, that's why this is equal to 0. This power, which I have to use 1 half to this power, get the 1. Next, log 1 half of 2 equals. All right. Now, just a little bit more complicated. What power should I use for 1 half to get 2? Well, again, it's kind of easy for me since this is less than 0 and this is greater than 0. It means I have to invert it, so my power should be negative. And considering I'm basically talking about easy problems, it should be negative and integer. Well, my first try and very successful one would be minus 1. Well, indeed, if you will do this, what does it mean raise 1 half into the power of minus 1? It's basically an inversion. From 1 half, I'll get 2. Exactly what I have to have here. So, 1 half raised to the power of minus 1 gives you 2. This is exactly the same thing. That's why this is a correct answer. Okay. Next, log base 1 half as usually of 1 eighths. Well, let's think about what power should I use to raise 1 half into to get 1 eighths. Well, the power of 1 gives me 1 half, power of 2 gives me 1 half times 1 half. So, it's 1 quarter. And finally, power of 3 gives me 1 half times 1 half times 1 half and that will be 1 eighths actually. So, it should be 3. And this is the verification. 1 half to the power of 3 is equal to 1 eighths. Oh, I'm sorry. 1 eighths. Okay. Log base 1 half of 8. Well, let's think about it. If using the power of 3, I get 1 eighths, then using the power of minus 3, I should get an inverse of 1 eighths, right? And indeed, 1 half minus 3 is equal to minus means it should be inverted. So, 1 half becomes 2, power of 3, 2 times 2 times 2 would be 8. So, 1 half raised to the power of minus 3 gives me 8. Now, again, let me emphasize, I guessed it. I mean, in this case, maybe not my first guess, but one of the first guesses was actually a successful guess. Obviously, this would not work for something more complicated. But for these little exercises, that's exactly what's necessary to do. You have to just approximate what might be the answer, positive or negative versus what exactly the relationship between these two base and expression under log everything. If they are less than 1 and greater than 1, then the power should be negative. If they are the same, both less than 1 or both greater than 1, then the power should be positive. Finally, the last one in this series, log 1 half of 1 sixth fourth sixth, right? Because 1 half of the sixth degree would be 1 sixth fourth, because it's 2 times 2 times 2 times 2 times 2 times 2. It's 2 times 6 times by itself in the denominator and that's 1 sixth fourth. Okay, that's the last one in this series. Just takes a little guessing as you saw. With log reasons, as I was saying, you have to do a little guessing in these simple cases. You cannot directly calculate like if you have some integer exponent and you can basically multiply things directly. And now I have the last series of the problems, which is exactly the same as the previous one, but now the base is 25. So base is 25 of 1 fifths. Okay, as I was saying before, this thing is greater than 1. This thing is less than 1. So most likely, I mean definitely, not most likely, definitely our exponent, the power we have to raise 25 to should be negative. So 25 becomes 1 25th. Now, how to get from 1 25th to 1 fifths? Well, you have to really extract this square root, right? Square root of 25, which is the power of 1 half. So my guess is it's minus 1 half. Let's just think about it. 1 to 5, if you will raise it into 1 minus 1 half, what does it mean? First, we have to invert it because it's minus, so it's 1 25th. And 1 half means square root. Square root of 1 25th is 1 fifths. Exactly what we have to get here. So 1 to 5 raised to the power of minus 1 half is 1 fifths. That's why this is the correct answer. Next, log base 25 of 625. Well, this is obviously 2 because 25 square is 625. All right, log base 25 of 5. Well, obviously this is 1 half because 25 half, which is square root of 25, is 5. Next, log 25 of 1 25th. What power should I use to convert 25 into 1 25th? Obviously minus 1 because 25 in the power of minus 1 is 1 25th. Right? Two more. Two more. As you see, these are all simple problems and they're basically are nothing but repetition of the definition of the logarithms. So if you know the definition of the logarithm, that this is the power you have to use to raise the base to get the expression under the logarithm. If you know this, every one of these problems is trivial. Okay, have two more. Log 25 of 1 625. Okay, you remember that log of 625 was 2 because 25 to the power of 2 was 625. If I want 1 over 625, it should be minus 2, right? Because we have to invert. And indeed, 25 to the power of minus 2 is 1 over 125th square, which is 1 over 625. And the last one, which I'm sure everybody considers as an easy thing, the logarithm of 1 base 25 or base 1 half or base anything would always be 0 because any base to the base to the power of 0 will give 1. So this was the last problem. And again, the whole lecture is short and simple. And its only purpose is to basically inculcate into your minds the definition of the logarithm. It's just a different notation for exponent. If you know this, you know that. Nothing new about this. With calculations of the value of the logarithm, here situation is slightly different because in simple cases, calculation of the exponent is actually simple. Calculation of logarithms involves certain guessing. But that's only in simple cases. In practice, it doesn't really matter because people don't really do anything right now outside of the computers. But in these simple cases, when integers are involved, you really have to understand that the guessing is really fine in this particular case. All right. So I do encourage you to use Unisor.com as a complete course, actually, which introduces you to different aspects of mathematics. And it would really help development of your analytical thinking, your creativity, etc. So do consider registering under a certain supervision. Get enrollment in some course, like algebra or geometry or whatever, because then you will be presented with an opportunity to go through exams. Exams are important. Here I am explaining this and it's kind of easy for you to follow. But if you have to really do the exam yourself, it really trains your mind relatively well. Thank you very much.