Description
For this exponents portfolio we learned about different types of exponents. We had to set it up into three different sections which was, Exponents rules, Exponential Growth and Decay models, and Forms of Exponential Equations.
Exponents Rules
For the exponent rules portion I chose the papers Exponents Review and Evaluating Exponential Equations.
Exponents ReviewOn the paper Exponential Review we started with the basics. On activity 4 part c, all you have to do is subtract the exponents then you get the answer. For example one of the problems is x^20/x^13. How you would solve this is you would subtract 20 and 13 which are the exponents. That equals 7 so then it would be x^7. In activity 5 it wants us to evaluate the expressions (7,654,321)^0 and 3^0+x^0+(3y)^0 if A^0=1. Since every number with the exponent of 0 would equal 1 then that means that (7,654,321)^0 would equal 1 and 3^0+x^0+(3y)^0 would equal 3. It would equal three because as you see in the equation there are three exponents of 0 so then its just 3.
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Evaluating Exponential EquationsFor this paper we got learned about two different equations. There are two characters, Henry and Henrietta, and they have two different ways to prove that 4^3/2=8. Henry's way of proving this is by turning it into a square root. For example he did this problem by turning 4^3/2 into (4^3)^1/2. Then this equals 64^1/2 which just means you square root it. 64 square rooted does equal 8 so his way does work. Henrietta did this problem by simplifying it. She turned 4^3/2 into (4^1/2)^3. After that she solved for the parenthesis first which means (4^1/2) turns into 2 then we just keep the exponent there. That means that it turns into 2^3 which once its solved is also 8.
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Exponential Growth and Decay
For this section I chose Population and food supply and Compounding with 5% interest.
Population and Food Supply
For this problem we had to figure out functions for things such as the population increasing or the food rate increasing with the information given. The problem stated that a countries population initially had 2 million people and was increasing at 4% per year. The food supply was initially good for 4 million people and is increasing at a constant rate of .5 million people per year. After they give us the information they then want us to find the functions for the population and food increasing. What I did was plug in the information for the population increasing and what I got was P(t)= 2 mil(1+.o4)^t. The function for Food supply increase would be F(t)= 4t0.5t. 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).
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Compounding with 5% Interest
For this packet we were given an equation. The equation was A=P(1+r/n)^nt. A stood for the amount in the account, P stood for the initial amount, r stood for rate(annual interest rate)written as a decimal, n stood for the number of compounds in a year and lastly, t stood for the time of investment in years. After learning about this equation we got to put it to use with practice problems. The practice problem 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 into the equation. After I plugged it in I got 30,000(1+.08/4)^4x10. When we simplify this it is 30,000(1.02)^40. When I solved it I got 66241.18991.
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Forms of Exponential Equations
Forms of Exponential ExpressionsFor 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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