Description
In this portfolio we learned about different types of exponents. I broke them up into 3 different sections, exponent rules, exponential growth/decay models and forms of exponential equations.
Exponent Rules
For the exponent rules section the two papers I chose was Exponents Review and Evaluating Exponential Equations.
Exponential ReviewOn the paper Exponential Review we started with the basics such as what 3x3 would be in exponential form, which would be 3^2 which also would equal 9. We then got more into unfamiliar territories such a evaluating different types of exponents. In activity 4 of this paper they gave the equation 2^7/2^3. You had to basically simplify it into a smaller exponent. 2^7/2^3 is equal to 128/8. That can also be simplified to 16 when u divide 128 and 8. Also 16 can be simplified to 2^4 and that is what I did for this section.
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Evaluating Exponential EquationsFor this packet we learned about two different ways to solve exponential equations. In the problem we had to figure out if these two people, Henry and Henrietta, were right about their methods. They both agree that 4^3/2 equals 8. Henry method is (4^3)^1/2 equals 64^1/2. Whenever we have a number written like 64^1/2 then that just means the square root of that number so that means that we are just finding the square root of 64. The square root of 64 is 8 so that means that henry was right. Now Henrietta's way is (4^1/2)^3 equals 2^3 because (4^1/2) just means 1/2 of 4 which is 2. Henrietta was also right because 2^3 equals 8. both of these methods work and both of them were correct.
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Exponential Growth And Decay
For the exponents growth and decay I chose Compounding with 5% Interest and Population and food supply.
Compounding with 5% Interest
In this paper we learned about and equation which was A=p(1+r/n)^nt. A stood for the amount in the account. P is the initial amount. R is the rate( annual interest rate)that is written as a decimal. N is the number of compounds in a year and t is the time of investment in years. One of the problems we got was " An initial deposit of $30,000 is placed in an account that earns 8% interest. Find the amount in the account after 10 years if the interest is compounded quarterly. What I did for this problem is plug it in to the given equation which is A=p(1+r/n)^nt. The equation would be 30,000(1+.08/4)^4x10. First I figured out what was in the parentheses then I did the exponents and after that I multiplied it by 30,000. The answer I got was 66,241.18991.
Population and Food Supply
For this problem we had to find different functions for the population and food supply. What I did was plug in the information that was given. The problem gave me an initial population of 2 million that increases at a rate of 4% per year. It then gave me the initial food supply amount which is good for about 4 million people that increased at a constant rate of .5 million people per year. The function for population would be P(t)=2mil(1+.04)^t. The function of food is F(t)=4t.5(t). The problem was then changed to the food supply to being doubled initially so the new function for the food supply was h(t)=8tx.5(t).
Forms of Exponential Equations
Forms of Exponential Expressions
For this problem we had to figure out if certain equations were equivalent to each other. What I did for these problems is break them down and solve them the most I could. I started by writing them down the I slowly stated to solve them. The function they give us is R(d)=7.35(1/2)^d/2. It then gives us two equations and asks us which one is equivalent to the first equation. The options were R(d)=7.35(0.250)^d and R(d)=(0.707)^d. What I did to decide of they are equivalent is plug in a number for d. I chose the number 2. When I plug 2 for the equation R(d)=7.35(0.25)^d then it turns into 7.35(0.250)^2 which equals .459375. The other equation would be 7.35(0.707)^2 and it would equal 3.674. Now back to the first equation R(d)=7.35(1/2)^d/2 with 2 plugged in would equal 3.675. We can see that obviously one of them doesn't add up right away which is the equation 7.35(0.250)^d. So now we know that the one that goes is 7.35(0.707)^d.
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