Question

Help please

Answer 1

any restrictions on x? keep in mind this is a cube root not a square root function

Answer 2

Um 0

Answer 3

the cube root can be taken for any x-value, positive or negative (cube root of 0 is still defined) so no restrictions on x

Answer 4

for 6: what value of x makes the log undefined?

Answer 5

5

Answer 6

good, since x = 5 makes the inside of the log 0 it's undefined there

for 7: any restrictions on x?

Answer 7

um 7?

Answer 8

does putting x = 7 make the function undefined?

Answer 9

no

Answer 10

would it be maybe a negative number

Answer 11

idk I'm having trouble with this one

Answer 12

good

e^x doesn't have any restrictions on its domain (x can be positive or negative)
so e^(x+4) + 7 doesn't have any restrictions either

Answer 13

that's it for 5,6,7

Answer 14

so no restrictions ofr 7

Answer 15

yup that's correct

Answer 16

also change #6 to x cannot be 5 ~~~or less~~~ since logs need to be positive

Answer 17

oh so i would not put anything for the restriction and just write x cannot be 5 or less since logs need to be positive

Answer 18

or do i put 5 as the restriction and then add this explanation

Answer 19

x cannot be 5 or less

Answer 20

Got it !

Answer 21

"x < 5" "x must be greater than 5 because the input/argument for a log must be positive"

Answer 22

okay and for this

Answer 23

8A Part I 3^5=x^2+18
Part II x=15,-15

Answer 24

yup good

Answer 25

B Part I 10^2=3x+4
Part II x=32

Answer 26

good

Answer 27

I'm having a hard time with C PArt I

Answer 28

first use the exponent rules to combine the e's into one e expression

Answer 29

e^x^2+1?

Answer 30

the rule is to add the exponents not multiply them

Answer 31

2x?

Answer 32

what is x + x + 1?

Answer 33

3x

Answer 34

combine like terms

Answer 35

oh shoot I'm sorry

Answer 36

2x+1

Answer 37

2x+1

Answer 38

good so e^(2x+1) = 1 is your soln' for part I

to convert this to a log for part II take the natural log of both sides

Answer 39

I dont even know what i was doing there :/

Answer 40

x=-0.5

Answer 41

good that's part III but for part II they only want the log expression so 2x + 1 = ln(1)

is your soln' for part II

Answer 42

loge(1)=2x+1

Answer 43

oh okay

Answer 44

yeah that works too, loge is the same as ln

Answer 45

i would just solve for n right

Answer 46

and substitute 5 for n

Answer 47

hold up a sec

Answer 48

judging by the equation, the first term is 27, and the common ratio is 0.1

so start with 27, multiply by 0.1 to get 2.7, then just keep multiplying by 0.1 until you have a list of the first five terms

Answer 49

27,2.7,0.27,0.027,0.0027

Answer 50

good, now we just use the geometric sum formula to find the sum of the first ten terms

Answer 51

29.9

Answer 52

good

Answer 53

okay

Answer 54

for 10 part I it's just asking you to re-write the original equation with 300 instead of T(x)

Answer 55

300=1500e^-0.4x

Answer 56

yeah just like that

for part II it just wants you to solve for x

divide both sides by 1500 and then take the natural log of both sides, then divide by -0.4

Answer 57

loge(300)=-0.4x

Answer 58

300/1500 = ?

Answer 59

0.2

Answer 60

good so ln(0.2) = -0.04x solve for x

Answer 61

40.24

Answer 62

good so about 40 years 3 months ish

Answer 63

for parts I/II you just need to plug in t = 0 and t = 15 into the equation, respectively

Answer 64

212

Answer 65

and

Answer 66

330.39

Answer 67

awesome

for part III just set T(t) = 125 and solve for t

Answer 68

-23.17

Answer 69

good (you probably forgot the negative sign on 0.04)

t = 23 (needs to be positive since this is time)

Answer 70

anyway I should probably go I have some stuff to take care of