1. Pre-Calc Lesson 5.1 Exponents -- Growth and Decay: Integral Exponents Laws of Exponents: 1.b x b y = b x+y 2. b x = b x-y 3. If b /= 0,1,or -1, b y then b x = b y iff x = y 4.(ab) x = a x b x 5. (a) x = a x (b) b x 6.If x ≠ 0, and a>0, and b>0, then a x = b x iff a = b 7.(b x ) y = b xy 8. b 0 = 19. b -x = 1 b x

2. Example 1: Simplify: (b 2 /a) -2 and (a 3 /b) -3 1 st – the negative on the exponent indicates taking the reciprocal of the base! soo-- (a/b 2 ) 2 and (b/a 3 ) 3 Now distribute the exponent on the outside of the parentheses a 1x2 ; b 1x3 b 2x2 a 3x3 voila -- a 2 ; b 3 b 4 a 9 Example 2: Simplify: (a 2 + b 2 ) -1 where a ≠ 0 and b ≠ 0 1 st – make the inside of the parentheses look like one big fraction. – (a 2 + b 2 ) -1 1 Now take the reciprocal of the whole thing. ( 1 ) 1 a 2 + b 2 Since the power is 1 this is not needed soo --- 1 a 2 + b 2

3. Example 3: Simplify: a) x 5 + x -2 and b) (x 5 )(x -2 ) x -3 x -3 a)x 5 + (1/x 2 ) (1/x 3 ) Now multiply top and bottom by the LCD -- x 3 x 3 (x 5 + (1/x 2 ) x 3 (1/x 3 ) x 8 + x = x 8 + x 1 b)Much different problem as there is no addition or subtraction -- x 5 + (-2) – (-3) = x 5 – 2 + 3 = x 6

4. Suppose that the cost of a hamburger has been increasing at the rate of 9% per year. Then each year the cost is 1.09 times the cost the previous year. Suppose that the cost now is $4. Some projected future costs would be: Time ( years from now) | 0 | 1 | 2 | 3 | ‘t’ | Cost (dollars) | 4 | 4(1.09) | ? | ? | ? | 4(1.09) 2 4(1.09) 3 4(1.09) t The above table suggest that the cost is a function of time ‘t’. Since the variable ‘t’ occurs as an exponent, the cost is said to be an exponential function of time ‘t’. C(t) = ??????? 4(1.09) t

5. Example 4: Using the cost function described earlier, find the cost of a hamburger: a) 5 years from now. 4(1.09) 5 = $6.15 b) 5 years ago. 4(1.09) - 5 = $2.60 Look at the graphs of the two functions at the top of page 170, can you tell which graph describes ‘exponential growth’ and which graph describes ‘exponential decay’ ?

6. Generally, all exponential growth or decay problems can be described by this model: A(t) = A 0 (1 + r) t  where A 0  the initial amount (or when ‘t’ = 0) r  the ‘growth’ rate (usually a percent) t  the time (usually in terms of years) Example 5: Suppose that a radioactive isotope decays so that the radioactivity present decreases 15% per day. If 40 g are present now, find the amount present: a) 6 days from nowb) 6 days ago A(6) = A(-6) = = 40(1 + (-.15)) 6 = 40(1 + (-.15)) – 6 = 15.09 = 106.06 Hw: pg 172 CE: #1-17 all; WE: #1-19 odd