Download
1 / 25
250 likes | 261 Views
Learn how exponential equations and functions work, apply them to real-life scenarios like bacterial growth and compound interest, and grasp the concept of exponential increases and decreases.
E N D
Lesson 115:Exponential Equations, Exponential Functions, Compound Interest
We know that 10 squared equals 100. Thus if we write 10 = 10 The left hand side of the equation equals 100. for the right hand side of the equation to equal 100, x must equal 2. 10 = 10 2 x 2 2
We will generalize this observation by saying that if two powers of the same base are equal, the exponents must be equal. If we write 10 = 10 The equals sign tells us that the expressions are equal. The bases are equal, so the exponents must be equal. 1.23 = x + 2 1.23 x + 2
Example: Solve 17 = 10 X + 2
Answer: Log of 17 = 1.23 x = -0.77
Exponential functions are important because they help us understand problems such as the growth of bacteria in biology, the voltage on a capacitor in engineering, radioactive decay in physics, and the growth of money in finance.
The equations on the left define two exponential functions. A sub t represents the amount of money, bacteria, electrons, etc. present at some time t.
The graphs of the equations show that when time equaled zero, the amount present was 200, and that the amount increased exponentially with time. These equations describe an exponential increase because the coefficients of t are 0.2 and 04. which are positive numbers. The graph with 0.4 goes up faster than the graph with 0.2 because 0.4 > 0.2. when t equals zero, the amount present is called “A of zero” and equals 200 for both of these functions.
If the exponential constant is a negative number, the graph of the curve goes down as t is increased.
These equations describe exponential decreases, because the amount present decreases as time increases. The graph of -0.4 goes down faster than the graph of -0.2 because the absolute vale of -0.4 is greater than that of -0.2.
When we write the general form of an exponential function, we use A of zero to represent the amount present when t equals zero and we use k to represent the coefficient of time in the exponent. If k is greater than zero (a positive number), the curve goes up. If k is less than zero (a negative number), the curve goes down.
Example: In the beginning, there were 400 bacteria in the dish. The number of bacteria increased exponentially. Three days later there were 2000 bacteria in the dish. How many bacteria would there be in 8 days?
Answer: A = A e 2000 = 400e 29,362 bacteria kt t 0 k3
Example: In the beginning there were 4.25 ounces of radioactive material. It decayed exponentially. Ten years later 3.92 ounces remained. How much would remain after 30 years?
Answer: 3.34 ounces
James puts $100 in a bank account at 8 percent interest compounded yearly. At the end of the first year, his account had accrued $8 interest. $100 x 0.08 = $8 Thus the total amount in his account was $108. this is 1.08 times the amount he deposited. Total amount = $100(1.08) 1
The next year he was paid 8 percent interest on $108, so the total amount at the end of the second year was Total amount = [$100(1.08)](1.08) = $100(1.08) 2 2
At the end of the third year, he would have total amount = $100(1.08) And at the end of n years, he would have total amount = 100(1.08) 3 3 n n
Following this reasoning, we can see that the total amount in a bank account at the end of n years can be written as Total amount = P(1 + r) Where P is the principal, or the amount deposited, r is the rate of interest, and n is the number of years. n n
In this problem the interest was compounded yearly. Advanced math books will develop the equation for the amount present if the interest is compounded continuously. This is the exponential equation for compounded continuously. A = Pe rt t
This function describes an exponential increase where A sub 0 is equal to P and r is equal to the rate, which is a positive number. The development of this equation for continuously compounded interest is beyond the scope of this book.
Example: Roger deposited $5000 in an interest-bearing account that paid 7 ½ percent interest compounded continuously. How much money did he have in 10 years?
Answer: $10,600
