hi class let's continue our discussion of exponentials and logarithms so i want to talk to you today about exponential equations and logarithmic equations so of course it goes without saying that you know we can't you know there's no point in doing all the things we've done so far with exponentials and logarithms unless uh we're willing to go all the way and deploy these rules to actually solve equations so indeed that's what we're going to do today so let's talk about exponential equations so we have a rule that basically tells us that if i have base i have a number raised to the x power and the same number raised to the y power then in other words what we're basically saying is if these two are equal to each other then that implies that x equals y okay so this is kind of an exponential equation type of corollary in some respects so what are my guidelines for actually so using that information we can actually solve exponential equations so we're going to basically isolate the exponential expression take the logarithm using the l and using the law of logarithms in order to basically isolate the variable and then solve for that variable so i know that's a lot of words so let's actually try some and see if we can do this so suppose we have this relationship here so what we're going to do is we're going to take the log of both sides and when we do that we're going to use the power rule to bring this x plus 2 out front and we're going to multiply this by log of 3 and this is going to equal to log of 7 and remember log of 3 and log of 7 are numbers okay those are not variables those are numbers okay so now i can subtract log of 3 from both sides so let me kind of do this right up here actually before i can do that i have to distribute my log of 3. so here's what i get here so now i get log of 3 times x plus 2 log of 3 equals log of 7 and now let's subtract 2 log of 3. so that gives me log of 3x equals 2 log of 7 minus 2 log of 3 and then finally divide by log of 3. so my final result and i think i have this on the next slide is going to be x is equal to log of 7 divided by log of 3 minus 2. okay so um in essence that becomes my solution okay and they wrote there's a little bit differently than the way i wrote mine okay but you can see we get pretty much the same result i wrote log of 7 minus 2 log of 3 divided by log of 3 and what they did is they just broke this into 2 fractions which is log of 7 over log of 3 minus two okay so they're exactly the same okay and if you do the calculation you'll see that we get exactly the same answer okay now i can check this answer to make sure it works okay just so just like whether it's a linear or a quadratic or any other type of equation i want to check my work i i have to make sure that it works so again it's going to be all about precision okay and this is where logarithms can be you know very you know very temperamental okay but if i take point two two eight uh five seven seven five six and i change the sign on that and add that to two okay and then i want to take uh 3 raised to this power so 1771 244 so let's remember that number so 3 raised to the 1.771 two four whoops uh one point seven seven one three two four and that's gonna give me you can see here seven but it gives me a little bit more right so again why is it off okay well because i rounded right so rounding the result is going to affect my answer okay this is just that's just the way it is okay so just expect to have some roundoff error in this okay that's that's one of the dilemmas that we face when we work with logarithms so what about this one well now what i'm going to do is i'm going to deploy a process that we talked about a few a few uh months ago called substitution and i'm going to let u equal to e to the x and if i do that i can make this substitution right here and i have u squared minus u minus six equals zero and now i can reverse foil this and i'll give me u minus uh six and excuse me it's not going to be six what am i saying u minus 3 and u plus 2 equals 0. so therefore u equals 3 and u equals negative 2 but that's not the answer i have to plug back in this exponential so e to the x equals three and e to the x equals negative two but this is not possible right okay i can't take the exponential of anything and get a negative answer okay that's not there is no number that allows me to do that so this becomes my only valid solution so now what do i do i take the ln of both sides so ln of e to the x equals ln of 3 and therefore x equals to the ln of three remember the ln of e is one right and i bring the x out front okay so if i try this on my calculator okay you can try this to make sure it works first let's take the ln of e and you can see i get exactly one as i expect okay and then if i take um ln of three so if i take e and then i raise that to the three and i take the ln of that result you can see i get exactly three back out which is what should happen okay so therefore here's my answer here okay and then as i point out to you before okay e to the x equals negative 2 has no solution right because i know the e to the x okay looks something like this right it will always be positive it can't be negative okay so this result is not allowed now i can also apply the same sorts of rules to logarithmic equations too okay and use the same sorts of ideas so i know that if i have log base a of x and i set that equal to log base a of y then that implies that x equals y okay so just like with logarithms it works the same thing with exponentials so my process is going to be pretty much the same i'm going to use the laws of exponentials and the laws of logarithms to isolate x and solve so let's consider this example here suppose i have ln of x equals eight so what do i have to do i have to take base e of both sides of this right so e to the ln of x okay will just become x and therefore this answer will be e to the eighth okay which is roughly 29 81. okay what about this one here let me do this one right down here let me let me write it down here so i've got log base 2 of 25 minus x equals 3. so log base 2 of 25 minus x equals 3. so now what am i going to do i'm going to do base 2 on both of these numbers so 2 to the log base 2 of 25 minus x is just going to be 25 minus x and then 2 to the third is just going to be 8. okay so now add x to both sides so let me do that right here and that gives me 25 equals 8 plus x subtract 8 from both sides so x equals 17. okay and you can see that's exactly what they got right here now remember to always check your work to make sure it's going to work so if i take um log base 2 of 25 minus 17 i get log base 2 of 8. so in other words when i ask myself the question log base 2 of 8 what am i asking what is the number such that i take 2 to that number and the answer is 8 well we know that that answer has to be 3. so therefore x equals 3 and this answer is 3. okay whoops sorry about that okay so um as i mentioned earlier on i mean logarithms have many applications in the real world one of the common applications comes with the refraction of light i don't know if you know this or not but when you shine a bright light through glass okay the light is pretty much going to not refract it's going to pretty much travel in the same direction and reach our eye in exactly the same trajectory that the light was shined right however okay if i take the same light and i shine it into water or i shine it into water with with that's got pollutants in it or something like that the water is going to refract so basically if this is the water level right here and these are the rays of light then what happens is that when it hits the water it's going to basically refract okay refracting means that it changes direction okay and what does that mean well if i'm down here in the water somewhere okay here i am right here okay and there's a big bright light okay and if i'm out if i'm out of the water if i'm standing up here someplace on a platform somewhere okay the light is just gonna wash all over me right but if i'm down in the water and the light shines on me then obviously my eye is not going to see the light in the same way okay and the reason why is because the light is being refracted it's being bounced in different directions it's being absorbed in some cases okay it's the the the the speed at which the light and the direction at which the light is traveling is going to be affected by the medium of water that's called refraction so suppose we have this example here if i zero and i denote the intensity of light before and after going through a material x is the distance in feet the light travels in the material and then according to the beer lambert law we have this following relationship where k is a constant depending on the type of material so it turns out that different materials have different refractive coefficients okay and you can actually look these up in in science books okay uh ice water um you know glass uh you name it okay any any material that is that is translucent okay in other words light some measure of light can travel through it okay it's gonna have a what's called a refractive index so this k value is like a for a refractive index essentially so we want to solve for i and then for a certain lake k is 0.025 and the light intensity is 14 lumens find the light intensity at a depth of 20 feet so let me make note of this equation okay so we have minus one over k and then we have ln of i over i naught and that's going to equal to x okay and let me check that to make sure yes i did okay so now i want to solve for i so the first thing i'm going to do is i'm going to move this coefficient up to the exponent so i'm going to write this as ln of i over i naught and all of this is going to be raised to the 1 over the negative 1 over k power and then i'm going to write an x here okay so now i'm going to raise both sides to base e so i'm going to take the base e of both sides okay so that transforms this equation to i over i naught and all of this is going to be raised to the 1 over k power and then this is going to equal to e to the x okay and remember x is is going to be the depth right x is the depth okay and now i'm going to raise both sides okay to the minus k power so i'm going to raise this side and this side to the minus k power because i'm trying to get rid of that fractional exponent so i over i naught is going to equal to e to the x raised to the negative k power which is just going to be e to the minus k x okay and then finally i'm going to add i'm going to multiply by i naught on both sides so my final equation let me write it right up here is i is equal to i naught e to the minus kx okay and now they tell me that the initial intensity is 14 and then of course we know e is a number and we know k is negative .025 and then they tell me that x is 20 okay so let me write 20 right there and now i have all the ingredients to solve for i so let's do that really quick so .025 change sign times 20 gives me negative 0.5 so e raised to the negative 0.5 and then multiply that by 14. and that gives me intensity of 8.49 so what's happened here okay the intensity of the light was 14 lumens when i shined the light on the lake okay with a you know a k or a refractive index of 0.025 okay and i'm basically 20 feet under water so i'm talking about 20 feet of water between between where the light is and where i am right so basically what's happened is that only only about 8.49 lumens of the intensity of the light is actually getting to my eye so yes i'll be able to see the light but it will be a lot dimmer okay it's not going to be nearly as bright okay so here is the derivation for the equation just to confirm that we got what we were supposed to get and then of course here's my result right here okay so as we would expect okay the light traveling through the lake water 20 feet okay is going to reduce the intensity significantly okay and i expect this right even with clear water okay the purest most clear crisp water you can think about there's still going to be a reduction in the luminosity of the light the intensity of the light okay why because the water is going to reflect the refract the light okay and perhaps maybe you've seen that if you take like a bat or something and you stick one into the bat into the water doesn't it look like the bat is kind of broken has anyone ever done that before you know what i'm talking about if i'm looking down on a lake and i have a stick or something and i stick the stick in the water okay when i look in the water the stick looks like it's kind of broken and separated has anyone ever noticed that okay why is that because of the refractive nature of the water the light that bounces off the object is hitting my eye and making it appear as though the stick is broken when of course it's not okay but it's just the refractive nature of the water that's why i'm seeing what i'm seeing so just like before we talked about compound interest and we talked about with exponentials how compound interest can be so useful to me okay well with logarithms the same thing can be said so let's take a look at this example here suppose we have a sum of five thousand dollars it's invested in an interest rate of five percent per year find the time required for the money to double if the interest is compounded according to the following methods okay so now what we're doing is we're basically in a position to where we have to use the rules of logarithms in order to be able to solve this because why because they're asking me to solve for the time and the time is in the exponent so we have five thousand dollars okay that's our p0 and we want the value to double right so that means my ending amount is going to be ten thousand okay so we have ten thousand equals to five thousand and then we have one plus and they give me the rate of .05 okay and the n varies right so they want it semi-annually and they want it continuously so let me go ahead and just do it semi-annually because the continuous equation is going to be completely different so let me get rid of this n value here and i'm going to change this to 2 okay for semi-annual okay and then 2 to the t right so the only hope i have of solving this is going to be logarithms so let's divide by 5000 and that gives me 2 equals 2 and let's go ahead and simplify this equation in here so 0.05 divided by 2 plus 1 gives me 1.025 1.025 and then 2t so now my only choice here is take the logarithm of both sides so let's do the log of 2 and the log of this exponential that you see right here okay but remember i can bring the exponent down and put it as a coefficient so 2t times the log of 1.025 and now if i solve for t let me do it right here t is going to equal to the ln of 2 okay divided by 2 times the ln of 1.025 so therefore t is going to equal to let's do this calculation 2 ln divided by 1.025 ln okay and then divide that by two so therefore how much time is it going to take 14 years okay 14 years approximately for the amount to double okay and that's not too bad okay ten thousand dollars after 14 years you know that's not too bad okay one of the reasons why it's growing so slowly okay is because it's only compounding semiannually okay if i want to get a much better rate of return then i want to look at something where it compounds more frequently so let's actually take a look and see what happens when it compounds continuously and i'm going to open a word document to do this here we go okay so remember to compound continuously p is equal to p naught e to the rt right and of course my initial amount is 5000 my terminal amount is 10 000 and then e and then .05 right that's the rate times time so the formula is completely different so make sure that you're mindful of the differences again divide by 5000 so i get 2 equals to e to the 0.05 t and now again take the ln of both sides so let me do that right here ln of 2 and then ln of e to the 0.05 t but remember i can bring that co that exponent down front and treat it as a coefficient so 0.05 t and then ln of e and we know ln of e is 1 right so this becomes ln of 2 equals 2.05 t so t is equal to ln of 2 divided by 0.05 so now let's see what the what the equivalent time will be if i compound continuously so 2 ln divided by 0.05 okay and that gives me 13.86 years okay so 13.86 so again not much of an improvement a little bit of an improvement so in any any conceptions that i would have about the the rate at which i'm you know earning doubling my money okay taking me 14 years in which to do this okay any any concerns i would have as far as the amount of compounding periods is probably unfounded okay um but okay um maybe i need to look more at a better interest rate maybe i need a better product uh in which to invest my money because it looks like that uh in terms of of this return here it's not going to really give me that much of a benefit so 14.5 for the first one which agrees with what i got so we're going to double in 14 years and then if i compound continuously okay as you recall 13.86 so i did make a little bit of an improvement okay but not very much okay so again if it were me okay and i was wanting to basically uh harvest more money i would probably look for a product that was going to give me a higher interest rate okay thank you