Question

Please help me with the following questions. Thanks in advance!!

Answer 1

Explanation:

In choices a and b, the bases are positive numbers greater than 1, and so these are growth functions. In c and d, the bases are between 0 and 1, and thus these are decay functions.

In the second problem we have 3ln(x + 1). Rewrite this as ln(x + 1)^3.

We also have 9ln(x - 4). Rewrite this as ln(x - 4)^9.

Because of the + sign connecting ln(x + 1)^3 and ln(x - 4)^9, these two logs combine to form

ln [ (x + 1)^3 ] * (x - 4)^9 (the log of a product).

Now we have:

ln [ (x + 1)^3 ] * (x - 4)^9 - 4ln(x + 7), or:

[ (x + 1)^3 ] * (x - 4)^9

ln ------------------------------------

(x + 7)^9

Answer 2

Explanation:

Looking at the table

a) f(x) = 0.35(10)^x is a growth function. The rate of change is greater than one.

b) f(x) = 8(3.75)^x is a growth function. The rate of change is greater than one.

c) f(x) = 2.1(0.79)^x is a decay function. The rate of change is lesser than 1

d) f(x) = 4.3(0.28)^x is a decay function. The rate of change is lesser than 1

5) 3ln(x + 1) + 9(ln x - 4) -4ln( x +

7)

Recall

alnb = b^a. Therefore, the expression becomes

ln(x + 1) ^3 + ln(x -4x)^9 - ln(x + 7)^4

= ln [(x+1)^3 × l(x - 4)^9]/(x + 7)^4