(3 s -5 t 3 s 2 ) 2

1.x 0 =1 2.x a x b = x a+b 3.x a = x a-b xbxb 4.(x a ) b = x ab 5.x -a = 1/x a 6.x a/b = ( b √x) a

These same properties hold for irrational exponents. Simplify 1.(2 √5 ) √5 2.(2 √5 ) (2 √5 ) 3.2 √(3) √(3) - 2

Some exponential equations can be solved by re- writing each side with a common base. Solve. 1.9 x = x = 4√2

 The exponential function with base b is denoted by f(x) = b x. Note that b represents a number. The variable is in the exponent in an exponential function. Often in applications we will use f(x) = ab x. (b >0, b≠1)

Graph. 1.y = 2 x 2.y = 3 x 3.y = (1/2) x List the domain, range, intercepts and asymptotes. Decribe the end behavior and whether the function is increasing or decreasing. How do the functions differ? How are they alike?

Domainall reals Rangey>0 y-intercept(0,1) x-interceptsnone HAy = 0 They differ in how quickly they “grow”. The first two increase from left to right, the third decreases.

Now graph these. Describe the transformation. 1.y = -2 x 2.y = 2 -x (which of the previous graphs is this equivalent to?) 3.y = 2 x+3 4.y = 2 x - 2

For many applications it is convenient to use the irrational number e ≈ 2.718… Graph y = e x. This is derived from lim(1 + 1/x) x as x → ∞. More about that later….

How much money would you have after x years if you started with $100 and added $5 per year? Is this linear or exponential growth? What if you earned 5% interest per year? Is this linear or exponential growth? What if you earned 5% interested compounded quarterly?

Compound A = P(1 + r/n) nt Continuous A = Pe rt A balance Pprincipal Rinterest rate (as a decimal) ttime in years nnumber of times compounded per year

 In 2000 there were 125 deer in a wooded area. In 2010 there were 264 deer. Write an exponential function to model the yearly growth of the population. (Do you have enough information to decide whether this is a good model?)Predict when the population will reach 500 deer.  Answer: N = 125(1.078) t  about 18.5 years

Use model N = N 0 e kt (A = Pe rt is an example) *N 0 will always be positive, k can be pos or neg *N 0 will represent the starting value, k the relative growth rate, t time N is like y, t is like x You need to know or find N 0 and k

At the beginning of an experiment you have 1200 bacteria, 4 hours later you have Write a continuous exponential growth function to model the situation. k≈.1515, so N = 1200e.1515t

 The half-life of a substance is the time it takes for 50% of the sample to disintegrate. (It is a property of the substance, it does not depend on the initial amount.)  For example, iodine-131 has a half-life of 8 days. If you started with 2 grams of this substance, graph the amount remaining after 0, 8,16, and 24 days. Then write a continuous exponential decay function to model the situation.  Answer: N = 2 ● e t