 Hi, I'm Zor. Welcome to a new Zor education. I would like to spend some time solving problems related to exponential actions, primarily related to both certain properties which we have already learned before, and the graphs which I also spent some time describing. Let's just start from something which I used without the proof during one of my first lectures on exponential functions. I was actually referring to the fact that exponential function like this with greater than one base is growing and it's growing without any limits. It eventually overcome any number whatever we can come up with. So basically it grows to positive infinity like we just use these words although I don't really like to use infinity because it's not properly defined, but anyway everybody understands that the function grows limitlessly, so to speak. Okay, how can I prove it? Now I referred to something which seems to be rather obvious like this and I said well it's very easy to prove it by induction. Okay, so as our first problem let's just prove it by induction that this is a true statement where d is greater than zero. Now I put one plus d instead of a with d is greater than zero because a is greater than one, so it's easier to do it this way. So this is basically this function for some value d which is greater than zero and whatever the integer, whatever natural actually exponent is. So if I will prove that eventually this function for natural argument of exponent overcomes any given number then obviously for or rational or irrational it will be proven as well because we were talking about monotonic property of the function. So if I will be able to prove this then it's very easy to prove that for any given d and any given number I will always find the exponent n that this particular expression would be greater than that number. If I want to find number n in such a way that the result of this will be greater than some k, what is the n? Well obviously it's solvable n greater than k minus one divided by d, right? k minus one divided by d. So if I will choose this n for this given k and d then my exponent would be greater, my exponent would be greater than this number k because it's greater than this and this is greater than k. So I was using it without a rigorous proof so let's just do it by induction. Okay how can I prove this by induction? First if you remember I have to check if the formula in this case its inequality is true for some initial value of n. Well we are talking about natural n, right? So we can check it for n equals to one but what's interesting is for n equals to one on the left I will have one plus g and on the right I will have one plus g which is inequality not inequality. That's not good so let's just move forward since we are moving always to infinity we don't have to start with one, we can start with two. So how about two? One plus d square is equal to one plus 2d plus d square and one plus 2d is on the right. Now obviously this is greater than this because of this positive element. So for n is equal to two I have checked that this is true. Okay first n equals to true. The second step in the induction proof is assume that the formula is true for some n let's say equals to some k whatever. So one plus d to the power of k is greater than one plus kd. So assume for n is equal to k. Now three set n is equal to k plus one. The next one. Well let's just think about one plus g. Now to the power of n where n is equal to k plus one is this which is one plus g to the power of k times one plus g. If you have the same base multiplied by each other exponents are added together k and one would be k plus one. Now this by assumption is greater than one plus kd. I will take this without any change. Well let's open the parenthesis it will be one plus kd plus d plus kd square which is equal to one plus k plus one d plus kd square. Now this is greater let's just draw this one right. This is greater than this. So basically we started with this, did some transformations, used our assumption and we came to the same formula as before for n is equal to k plus one. You see here is k plus one and here is k plus one. k plus one k plus one. So the formula has been proven in the collage in this case has been proven for n is equal to k plus one. Considering we have checked for n equals two and assuming that it's true for n is equal to k that basically concludes our proof by induction. That was easy. Next. Next we are talking about steepness. Okay the function has certain quality which we can call steepness of this function. Let me just generalize it first. If you have some kind of graph whatever the graph is and you would like to know how steep the function is at certain point somewhere. Now you basically feel that this is less steep than let's say here because here we can put the tangent here and this tangent is here. This is steeper. So basically steepness is more or less a steepness of the tangent in this particular point whatever we are trying to measure. However since tangents to complicated functions is not an easy concept it goes to analysis limits etc etc we are not touching with this particular thing. So they would like to approximate the steepness of the function and here's my way to approximate. Instead of dealing with tangents I will deal with increments. So if this is two integer points of the argument I have if this is the function this particular segment can characterize how steep the function is. If moving by one step from n to n plus one my function is increasing from a to the power of n to a to the power of n plus one then their difference characterizes the steepness of the function. So on a unit increment of the argument function is incremented by certain number and the greater this number is the steeper the function is. So my point is that if I will have two different exponential functions this is b to the power of x. My point is that the greater the base if b is greater than a the function is steeper which means that on the same segment the increment of the function this is b to the n and this is b to the n plus one. So the increment of the function b to the power of x where b is greater than a should be greater than the increment of the function a to the power of x. If I will be able to prove it then I can state that the exponential function with larger base is steeper than the corresponding exponential function with a smaller base and we are talking about case when both bases are greater than one so the function is increasing not decreasing. So assuming that I have two different bases both greater than one a smaller b is larger I would like to prove this. So if the base is smaller then the increment on the same increment of the argument increment of the function on the same increment of the argument from n to n plus one would be larger for greater base. It kind of feels this way but how to prove it well actually the proof is very easy because what we can do is a to the power of n plus one is a to the power of n times a minus a to the power of n. Now b to the power of n plus one is b to the power of n times b minus b to the power of n. Right? Now this is a to the power of n factor out a minus one this is b to the power of n b minus one. Now both a and b are greater than one so these are positive also b is greater than a which means b to the power of n is greater than this and b minus one is greater than a minus one. So we have two smaller numbers and this is two larger numbers multiplied by itself. So obviously the inequality is this way and obviously everything is reversible that's basically the proof so we can get to this particular inequality. So the larger the base the steeper the function and we are talking about case when the base is greater than one. Okay now we have a different set of problems related to some calculations and properties of the exponent. Now right now I'm talking about only rational exponent and I would like to remind that a to the power of p over q by definition is q's root of a to the power of p where p and q are natural numbers integer greater than 0. Okay now considering this is a definition of the rational number and also by the way a to the power minus q over q would be one over that's the definition of negative exponent. Now using this I have a bunch of inequalities which I would like to prove. Okay two to the power three over two less than three. How can I prove this? Well let's go by definition. What is this? This is square root of two to the third and this is three. Now I don't know really whether this particular inequality is true or not but I can always square it in this case because this is square root if I will square it I will get rid of the square root. I will get two to the third three to the second. Now this is eight this is nine. Now this is obvious inequality right? So from something which I don't know of using certain transformations I came to the obviously true inequality. Now I'm saying the key statement since all the transformations are reversed reversible from this obviously true statement follows this. Now why is it reversible? Well from here to here is obvious. Now if I have two numbers positive numbers from both I can take an arithmetic root in this case it's a square root and that would not change the inequality. So from this follows this and this is actually a definition for this. So basically all transformations are reversible and that's why this is a proof. So analysis is going this way and proof is going that way. Now I'm not going actually to this exercise of explaining I'll just say since it's reversible I'm proving the top in the quality. So for all subsequent cases I will be as short as possible. Okay three to the power of three seconds should be greater than five. True or false? Okay this is square root of three to the third greater than five. Three to the third greater than five to the second right? I can square both sides of the inequality. Now this is twenty-seven twenty-seven and this is twenty-five. Now this is obviously true statement and all these are reversible that's why I have proven this one. Next five to the power of three seconds greater than eleven. Okay this is a square root of five to the third greater than eleven. Square both sides five to the third greater than eleven square. This is one twenty-five. Five times five times five twenty-five times five one twenty-five. Eleven times eleven is one twenty-one. Obvious inequality so from this everything is reversible follows this and this is the proof. Okay next three to the power to third greater than three. All right now this is the third degree root of three to the second greater than two that's the definition of three to the power two to third. Now I will make a erase both sides in the third power of three so I will get rid of the cubic root it will be three to the second greater than two to the third. This is nine this is eight. This is obvious everything is reversible and that's why this is proven. Next five to third less than three how to prove that. Again this is a cube root of five to the second raised to the power of three five to the second less than three to the three. This is twenty-five this is twenty-seven this is obvious and everything is reversible so my initial inequality is proven and the last one and the last one is I have to prove that eight to the power two-third is equal to four. Okay what is eight to the power of two-third this is the cubic root of eight to the second I have to prove this or if I will raise both sides to the power of three eight square is sixty-four four four cube is four times four sixteen times four sixty-four this is obvious and that's why everything is reversible and that proves this one. So this completes my short set of problems related to exponential functions. I will have another set of problems a little bit more difficult that will be in the next lecture. I do suggest you to go through all these exercises yourself go to unison.com website and everything is in there. Thank you very much.