REVIEW 7.1-7.4

Exponential Growth Models When a real-life quantity increases by a fixed percent each year, the amount y of the quantity after t years can be modeled by: y = a(1 + r)t a: initial amount r: percent increase (expressed as decimal) growth factor: 1 + r

COMPOUND INTEREST P = Initial Principal r = annual rate (as a decimal) n = times per year compounded t = number of years What do I put in for “n”? Quarterly → n = 4 Monthly → n = 12 Daily → n = 365 Yearly → n = 1

EXPONENTIAL DECAY MODEL y = a(1-r)t Decay Factor: 1-r a: initial amount r: percent decrease as a decimal t: time

EXPONENTIAL GRAPHS Exponential Growth Exponential Decay y = a•bx (b > 1) y = 3(2)x y = a•bx (0 < b < 1) y = 3(0.5)x

Continuously compounded interest

Definition of a Logarithm with base b A LOG IS ANOTHER WAY TO WRITE AN EXPONENT logby = x if and only if bx = y Logarithmic form: logby = x Exponential form: bx = y b and y are positive numbers b ≠ 0

Apply Properties of Logarithms Section 7.5

Properties of Logarithms Product Property: Quotient Property: Power Property:

Examples Expand each expression

Expand 6 log 5x3 y 6 log 5x 3 y = 5x 3 y 6 log – = 5 6 log x3 y – + = EXAMPLE 2 Expand a logarithmic expression Expand 6 log 5x3 y SOLUTION 6 log 5x 3 y = 5x 3 y 6 log – Quotient property = 5 6 log x3 y – + Product property = 5 6 log x y – + 3 Power property

The correct answer is D. Power property Product property EXAMPLE 3 Standardized Test Practice SOLUTION – log 9 + 3log2 log 3 = – log 9 + log 23 log 3 Power property = log (9 ) 23 – log 3 Product property = log 9 23 3 Quotient property = 24 log Simplify. The correct answer is D. ANSWER

Change of base formula If a and c are positive numbers with c ≠ 1 Then

EXAMPLE 4 Use the change-of-base formula 3 log 8 Evaluate using common logarithms and natural logarithms. SOLUTION Using common logarithms: = log 8 log 3 0.9031 0.4771 3 log 8 1.893 Using natural logarithms: = ln 8 ln 3 2.0794 1.0986 3 log 8 1.893