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Solving Formulas and Exponential Functions - Helping You Navigate Algebraic Equations and Exponential Growth Models

Learn how to solve formulas by using algebraic rules and solve exponential functions for various real-world applications. Explore compounded interest and exponential growth with step-by-step examples.

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Solving Formulas and Exponential Functions - Helping You Navigate Algebraic Equations and Exponential Growth Models

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  1. Solving formulas: When solving formulas, all variables will be given except one. To solve a formula is to use algebraic rules for solving equations to determine the value of the variable not given. Procedure: To solve a formula: Step 4: Write solution using units mentioned in problem. Next Slide In this section, we consider applications of exponential functions. Exponential functions describe a variety of naturally occurring things, like the decay of radioactive isotopes or the natural cooling of substances. They also model things such as population growth, compound interest, inflation and depreciation. This section is going try to make you aware of these instances and help you make some determinations of your own. Step 1: List variables given in: variable = constant form. Step 2: Substitute given values for variables in formula (rewrite formula using values given). Step 3: Solve for unknown variable by using algebraic rules for solving equations.

  2. Compounded Interest Formula So far, problems involving applications of interest have been limited to the use of the simple interest formula, I = Prt. In most cases, banks pay compound interest (interest paid on both principle and interest). The formula for compound interest is an important application of exponential functions. If ‘P’ dollars is deposited in an account paying an annual rate of interest ‘r’ compounded (paid) ‘n’ times per year, the account will contain dollars after ‘t’ years. In the formula above, ‘r’ is expressed as a decimal and some values of ‘n’ are as follows: compounded semiannually: n = 2 compounded daily: n = 365 compounded annually: n = 1 compounded quarterly: n = 4 compounded monthly: n = 12 Next Slide

  3. A = ? P = $5000 r = .08 n = 4 t = 3 The account will contain $6,341.21. (The actual interest earned = $1,341.21) Your Turn Problem #1 Find the amount of money in an account after 7 years if $8500 is deposited at 5% annual interest compounded semiannually. Example 1. Find the amount of money in an account after 3 years if $5000 is deposited at 8% annual interest compounded quarterly. 1. Write down the compound interest formula and the given information. • Now, substitute the values • into the given equation. 3. Simplify the exponent. 4. Simplify inside the parentheses. 5. Raise 1.02 to the 12th power (use the yx or ^ key on the calculator), then multiply result by 5000. (Round to the nearest cent) Answer: $12,010.28

  4. Law of Exponential Growth The ideas behind “compounded continuously carry over to growth situations. We use the law of exponential growth, as a mathematical model for numerous growth-and-decay applications. In the equation, Q(t) represents the quantity of a given substance at any time t, Q0 is the initial amount of substance (when t = 0) and k is a constant that depends on the particular application. If k < 0, then Q(t) decreases as t increases. When k > 0, then Q(t) increases as time increases. Example 2. The number of grams of a radioactive iodine, I-131 present after t days is given by the equation Q(t) = 150e-0.085t. How many grams will be left after 7 days? Q(t) = ? t = 7 There will be 82.7 grams of I-131 after 7 days. Your Turn Problem #2 Suppose the number of bacteria present under certain conditions is given by the equation Q(t) = 200e0.15t, where t is expressed in minutes. How many bacteria are present at the end of 10 minutes? 1. Write down the formula and the given information. • Now, substitute the values • into the given equation. 3. Simplify the exponent. 4. Raise e to the –0.595 on the calculator), then multiply result by 150. (Round to the nearest tenth) Answer: 896 bacteria

  5. Q(12) = 8000 t = 12 Therefore, there were approximately 66 bacteria present initially. Your Turn Problem #3 Suppose the number of bacteria present in a certain culture after t hours is given by the equation Q(t) = Q0e0.03t, where Q0 represents the initial number of bacteria. If 520 bacteria are present after 4 hours, how many were present initially? Example 3. Suppose the number of bacteria present in a certain culture after t hours is given by the equation Q(t) = Q0e0.4t, where Q0 represents the initial number of bacteria. If 8,000 bacteria are present after 12 hours, how many were present initially? 1. Write down the formula and the given information. • Now, substitute the values • into the given equation. 3. Simplify the exponent. 4. Divide by e4.8 on both sides to solve for Q0. (Round to the nearest whole number) Answer: 461 bacteria

  6. Graphing Exponential Functions We graphed exponential functions in the previous sections. The only difference will be that the base is not a rational number. The base we will be using is ‘e’. Just remember to plot enough points to obtain an accurate drawing of the function. Solution: y axis ( 2, 9.4 ) Notice the graph does not go below the line y =2. This line is then an asymptote for this function. Given a function f(x)= ax+k, y=k is a horizontal asymptote. x y 0 3 1 5.7 ( 1 , 5.7 ) 2 9.4 -1 2.4 ( -1 , 2.4 ) -2 2.1 ( 0 , 3 ) ( -2 , 2.1 ) x axis Next Slide Choose the value for x which will give zero in the exponent. Then choose at least two values for x to the left of that number and at least two values to the right of that number.

  7. Your Turn Problem #4 y axis (2, 4.4) (1, -0.3) (-1, -2.6) (0, -2) x axis (-2, -2.8) The End B.R. 1-7-07

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