1) The regular price of a pair of jeans is x dollars. Let f (x) = x – 5; g (x) = 0.6x

a. Describe what functions f and g model in terms of the price of the jeans.

Functions f and g model the discounted price of the jeans. Function f(x) models that new price is 5 dollars less than its regular price, function g(x) models that new price is 0.6 times its regular price, meaning 40% less than the regular price.

b. Find (f with g) (x) and describe what this models in terms of the price of the jeans.

(fog)(x) = f(0.6x) = 0.6x – 5

This models that new price would be 0.6 times regular price, meaning 40% discount on regular price, and then additional discount of 5 dollars.

c. Repeat part (b) for ( G with f) (x)

(gof)(x) = g(x-5) = 0.6(x-5) = 0.6x – 3

This models that new price would be 0.6 times regular price, meaning 40% discount on regular price, and then additional discount of 3 dollars.

d. Which composite function models the greater discount on the jeans, f wit g or g with f? Explain.

Since both (fog) and (gof) give 40% discount on regular price, but since (fog) gives additional discount of 5 dollars while (gof) gives additional discount of only 3 dollars, (fog) models greater discount on jeans.

e. Find f to the “-1” power and describe what this models in terms of the price of the jeans.

f¯¹(x) = x + 5

This models that regular price of jeans is 5 dollars more than the discounted price.

Notes: (F with g ) Circle means “with” The composition of the function f with g is denoted by f “circle” g and is defined by the Equation (f “circle”g (x) =f(g(x) ) The domain of the composite function f”circle” g is the set of all x such that. 1x is in the domain of g and 2. g(x)is in the domain of f.

Next set of problems ii) Use properties of logarithms to condense each logarithmic expression. Write the expression as a single logarithm whose coefficient is 1.(NOTE: the numbers next to LOG are small numbers)

problem # 1 log x + 3 log y = log(xy³)

problem # 2 5 ln x - 2 ln y =ln

Problem # 3 ½ (log5 x + log5 y) - 2 log5 (x + 1) =

Problem #4 log4 (3x + 2) = 3

Ans.

3x + 2 = 4³ = 64

Or 3x = 62

Or x = 62/3 = 20.6667

problem # 5 5 ln12x² = 20

ln12x² = 20/5=4

or 12x²=e^4

or x²=(e^4)/12

or x = e²/sqrt(12) = 2.133

problem #6 log3(x – 5) + log3 (x + 3) = 2

log3 [(x-5)(x+3)] = 2

or (x-5)(x+3)=3²=9

or x²-2x-15=9

or x²-2x-24=0

or (x-6)(x+4)=0

So, x=6, -4

x=-4 is not possible, because values inside logarithms then become negative.

So, x = 6

iii An artifact originally had 16 grams of carbon-14 present. The Decay model A=16e to the -0.000121t power describes the amount of carbon -14 present after t years. Use this model to solve 11

11) The half-life of the radioactive element krypton-91 is10 seconds. If 16 grams of krypton-91 are initially present, how many grams are present after 10 seconds? 20 seconds? 30 seconds? 40 seconds? 50 seconds? All the work has to be shown

Let the equation for krypton-91 is A = 16

Since at t=10, amount is halved, A(10)= 8 grams. So,

8 = 16

Or = 0.5

Or -10k = ln 0.5 = -0.69315

Or k = 0.069315.

So model is A = 16*

Amount after 10 seconds = 16* = 8 grams

Amount after 20 seconds = 16* = 4 grams

Amount after 30 seconds = 16* = 2 grams

Amount after 40 seconds = 16* = 1 gram

Amount after 50 seconds = 16* = 0.5 gram