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Exponents & Logs. Growth/ Decay/ Change of Base Mr. Morrow 2/26/2013 – 3/1/2013. - Warm Up -. Prove that the given equation is an identity: Find the exact value of: Simplify the given expression:. - Homework Questions -. - Exponents -.
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Exponents & Logs Growth/ Decay/ Change of Base Mr. Morrow 2/26/2013 – 3/1/2013
- Warm Up - • Prove that the given equation is an identity: • Find the exact value of: • Simplify the given expression:
- Exponents - Below I have a table of the growth (increase) in my tuition bill at VCU taken in the Fall of each year. Fall 07-11 are undergraduate fees while Fall 12 jumps to graduate fees. Let’s find the average rate at which VCU increases their tuition and figure out how much you would pay if you went to VCU in the Fall of 2015. ×1.07 ×1.04 ×1.03 ×1.09 So we could estimate the cost of tuition by: We know though, that Fall 07 is the initialterm, 1.06 is our rate, and the exponent is # of years after the initial.
- Exponents - We know though, that Fall 07 is the initialterm, 1.06 is our rate, and the exponent is # of years after the initial. We see a formula begin to develop looking like: So now how much you would pay if you went to VCU in the Fall of 2015?
- Chapter 5 - • What are we looking to do here… • Define and apply integer exponents • Define and apply rational exponents • Define and use exponential functions • Define and apply the natural exponential function • Define and apply logarithms • Prove and apply laws of logarithms • Solve exponential equations and to change logs from one base to another Our real objective is to get reacquainted…??? with this information in order to apply it to arithmetic and geometric sequences and series which are linear and exponential functions.
- Chapter 5 - We saw a few slides ago that the cost of tuition is increasing, what is this called? Inflation How does inflation impact your life and your families lives? What are some examples of products that have experienced a lot of inflation and/ or deflation.
- Chapter 5 - So what does depreciation look like in real life? Suppose we purchase a brand new 2013 Scion FR-S for $25,750. But the car does not hold that value the whole time you own in, it actually depreciates at 12% each year What is the value of the car after: 1. 1 year? 2. 2 years? 3. 5 years? 4. t years?
- Chapter 5 - We can solve this problem the same way we did the inflation problem. We know the following: Obviously this should look similar to the VCU example: What is the value of the car after: 1. 1 year? 2. 2 years? 3. 5 years? 4. t years? What is the most significant difference you notice between the inflation equation and the depreciation equation? - Use your calculator to investigate
- Examples of Inflation/ Depreciation Graphs - Exponential Growth Exponential Decay
- Chapter 5: Equations - Growth and decay can be modeled by the following equation: where A(t)is the future amount A0 is the original amount (when t=0) r is the rate of growth or decay t is the number of time periods If the base if greater than 1 (r> 0) → exponential growth If the base if less than 1 (r< 0) → exponential decay What do you expect to happen if our base, A0, is multiplied by -1? Why??
- Practice - Complete the table:
- Practice - The CDC released the following table that shows the number of Flu Vaccinations distributed over a 5 year period. Determine the rate of growth and number of shots that they will need to prepare for in 2015. 3 sig figs: ×1.09 ×1.09 ×1.09 ×1.09 That’s a lot of…
- Law of Exponents (Review) - Same base case If b≠ 0, 1 or -1 then if and only if x = y.
- Law of Exponents (Practice) - Same base case
- Law of Exponents (Practice) - Same base case
- Law of Exponents (Review) - Not the same base case If x≠ 0, a > 0, and b > 0 then if and only if a = b Power of Powers
- Law of Exponents (Practice) - Same base case
- Law of Exponents (Practice) - Same base case
- Rational Exponents - Earlier we looked at the case of inflation on the price of tuition. -How would we figure out the cost half a year from now (Spring semester for instance) We would use rational exponents We also call these rational exponents as roots:
- Rational Exponent (Practice) - • Adding like radicals: • Homework: pg 178 (1-37 odd)
- Warm Up - Suppose and , where . Find: Simplify the given Expression: Solve the following equation for by using trigonometric identities. Check answer with calculator. SHOW WORK
- General form of Exponential Equations - At the start of this chapter we saw the general form of our exponential equation begin to come together with our VCU example. The actual general form we will use for the class looks like the following: Where is the amount after t periods is the initial amount (t = 0) ris the rate (1 ± the percentage) tis the amount of time (periods)
- Practice - The number of wolves in the wild, in the northern section of Cataragas county, is decreasing at the rate of 3.5% per year. Your environmental studies class has counted 80 wolves in the area. How many wolves do you predict to be in the area after 47 years? (growth or decay)? wolves
- Practice - Five years ago I went up to Mt. Tremblant in Southeastern Canada. The Tuesday we were up there I was caught in an blizzard, went off of the path in the trails and wound up stuck in the snow for 8 hours. Assuming a normal body temperature and a 12% reduction of body heat every hour, what can you tell me about the situation of my body temperature modeled by:
- Doubling and Half Life - Occasionally, there are cases of growth and decay when we know (or want to know) how long it will take for something to increase or decrease to a fixed amount. If we are doubling the amount of something then it is increasing by 200% for each t: If we are halving an amount then we are looking for 50% Both of these forms use a special variation of our equation: is the initial amount when t = 0 b is the growth/ decay rate (2 or 0.5) k is the time needed to by b t is the actual time
- Practice - A bacteria population of doubles every 8 hours. What will the population be in: 1. 8 hours? 2. 16 hours? 3. 24 hours? 4. t hours? Let use the equation we just saw since we are doubling: b = 2 (why?) k = 8 t = 8 This will yield the following equations:
- Practice - The amount of light energy we receive from the distant star Sirius doubles every 1300 years. If we began receiving light energy from this star in 4,487 BC at just 1 joule, how many units are we receiving today? joules
- Exponential Decay - Radioactive isotopes have many useful applications in medicine, forensics and geology (they decay at a known, fixed rate) We refer to the decay rate of an isotope as the ‘half-life’. This is the amount of time that must pass for half of the isotope to decay into a new element. Example: The half-life of Na-24 is 15 hours. What percent of the amount present now will remain 24 hours from now? We would have about 33% of the original amount after 24 hours.
- Exponential Decay - Determine the half-life of Li-12 if we can model the decay by the following equation: , thus our half-life is
- Cha-ching$ - A few of you have started working and possibly even began to save money… …probably not though However, when you do want to start earning money at a bank (interest), we would use the general formula of: Suppose you deposit a paycheck of $100 and it earns 12% annual interest. How much would you have after 1 year?
- Compounding - Well we saw how to use the formula we’ve been using for awhile. How can make more money using the same form of this equation? Let’s not just compound it once per year, let’s compound it several times per year. Our only new variable is n which represents the number of times that the function compounded during one period t
- Compounding - Suppose a bank is willing to offer a 12% interest on your $10 deposit. Which type of compounding would you rather have: Annual Semi-Annual Quarterly So we see the more this money is compounded the more money we have in the end. This is very important when saving and/or taking a loan…
- How n effects Future Value - ($100 invested for one year at 12% annual interest) 100(1.12)1 = 112 100(1.06)2 = 112.36 100(1.03)4 ~ 112.55 100(1.01)12 ~ 112.68
- Compounding - As the number of times we are paid interest (the compounding) increases, we can see a relationship (a pattern perhaps?) emerging: Yea not that kind of pattern…this kind: What happens as ‘n’ becomes very large? 2.5937 2.7048 2.7169 2.7181 2.7183
- Natural Exponent e - An Euler concept • The relationship as n becomes very large can also be expressed as : • This equation results in the irrational number that starts as 2.718281828… which is known as e or the natural exponent. This is your first exposure to the concept of limits. • This erepresents continuous compounding, which is a useful, practical and mathematical concept
- Continuous Compounding - In any situation where compounding is ‘continuous’ we can determine the future amount through the following equation: Where: A(t)is the future amount at time ‘t’ is the initial amount ris the annual rate (as a decimal) tis the number of years
- Practice - Suppose we deposit $2,500 in a bank account that offers a continuously compounding annual interest rate of 3.12%. What is the balance in the account after: 1. 1 Year 2. 3 year and 8 months 3. 4 months
- Exponential Equations - Growth and decay: Doubling, tripling, half-life: Compounding: Compound continuously: Homework: pg. 183 (3-13 odd) pg. 189 (1-11 odd)
- Warm Up - Simplify: Solve:
- Logarithms - • Just as we had Exponent Laws, we also have laws for logarithms known as “Log Laws” • If b > 0 and then logbx1 = logbx2 if and only if x1= x2 • Just as we did with exponents we can solve for x by equating the arguments if we have a common base • Example: log3(4x + 10) = log3(x + 1) therefore • 4x + 10 = x + 1 and we can solve for x • logbMN= logbM + logbN • logbMk = klogbM for any real number k
- Practice - Express logb MN2 in terms of logbM and logbN. • Solution: • logbMN2 = logbM + logbN2 • = logbM + 2logbN
- Practice - Express in terms of logbM and logbN Solution:
- Practice - Simplify log 45 – 2log 3 Solution:
- Practice - Express y in terms of x if Solution:
- Practice - Solve log2 x + log 2 (x – 2) = 3 Solution: log2 x + log 2 (x – 2) = 3 log2 x(x – 2) = 3 x(x – 2) = 8 x2 – 2x – 8 = 0 (x – 4) (x + 2) = 0 x = 4 or -2 eliminate -2 as you can’t have a negative answer