Question

HELP!!
suppose that G(x) = log (base 4) (2x+2)-2
a) what is the domain of G?
b) what is g(31)? what point is on the graph of G?
c) If G(x)=3, what is x? what point is on the graph of G?
d) what is the zero of G?

Answer 1

\[G(x)=\log_4(2x+2)-2\]
domain -> all values of "x" when G can be calculated..

Answer 2

so, when do you think you cannot calculate "G"?
hint: can you calculate log of "0" or any thing negative?

Answer 3

can not calculate anything that's below zero right??

Answer 4

"log" of "0" and negative numbers is not defined

Answer 5

so, we write,
\[2x+2>0\]

Answer 6

simplify that and what is the inequality for "x"?

Answer 7

-1

Answer 8

you write it as
x > -1

Answer 9

yeah just noticed

Answer 10

great.
so, the domain for G(x) is, any "x" greater than 1
i.e., from "1" ( but not equal to 1) to infinity

Answer 11

so, domain = \((1,\infty)\)

Answer 12

okay!

Answer 13

how would you find b? since i cant do log base 4 on calculator

Answer 14

yes
remember the change of base rule?
\[\log_a(x)=\frac{\log_{10}x}{\log_{10}a}\]

Answer 15

a=4
log (base10) you have it on your calcultor

Answer 16

we didn't learn that yet.

Answer 17

hmm. that is the way you can calculate logarithm to any base.

Answer 18

so it would be log base 10 31/ log base 10 4?

Answer 19

oh no wait.. my bad
I took the problem wrong.. and you didnot even correct me!!!
\[G(x)=\log_4(2x+x)-2\]
this is what you have, right?

Answer 20

right?

Answer 21

yes

Answer 22

ok, so doing the steps for (a) again
\[2x+x>0\]

Answer 23

are you sure there are no exponents in ther?

Answer 24

no exponents

Answer 25

ok.. so, proceeding the same way now, what is your domain?

Answer 26

what happens to the -2?

Answer 27

-2 is not in the paranthesis of the "log" so, it just shifts the function G(x) up,down along the Y-axis

Answer 28

wait im sorry it is actually right the first time I did it wrong on here! it is actually (2x+2)-2. I am SUPER SORRY!!!!!!!!

Answer 29

write it exactly using the equation editor (button is below this reply box)

Answer 30

\[G(x)=\log _{4}(2x+2)-2\]

Answer 31

so, by co-incidence, I got it right!!!

Answer 32

yes! sorry i missed typed it on here!

Answer 33

so would it be the same for b?

Answer 34

yes. would not change. replace "x" by 31 and evaluate

Answer 35

I got 2.48

Answer 36

does that look correct?

Answer 37

nope

Answer 38

2*31+2=?

Answer 39

44?

Answer 40

\(2\times31+2=?\)

Answer 41

64. so the base rule dosent work since I used it and got the 2.48

Answer 42

wait
now do
\[\log_4(64)\]

Answer 43

using base-change rule..
it always works as long as you change to valid bases

Answer 44

oh okay so now it would be \[\left(\begin{matrix}\log _{10}64 \\ \log _{10}4\end{matrix}\right)\]

Answer 45

perfect.. (if you mean division)

Answer 46

hahaha yes

Answer 47

so the answer would be three!

Answer 48

makes more sense now

Answer 49

great,
now finally,
G(31)=3-2

Answer 50

will be 1

Answer 51

bingo yay..

Answer 52

so i am super confused on how to do the last 2 problems

Answer 53

how would you start c?

Answer 54

wait..
with (b).. what is the point?

Answer 55

you put it as (x-value,y-value)

so the point is (31,1)

Answer 56

point is (31, 1)

Answer 57

yes.
for G(3)
put x=3 and evaluate..
show me your steps so I know what you are doing

Answer 58

so it would be \[3=\log _{4}(2x+2)-2\]

Answer 59

close... you still have the "x"..
rewrite the given function as it is with all "x" replaced by "3"

Answer 60

including the "G" part

Answer 61

I thought that it was g(x)=3

Answer 62

nonono ..
just the "x"

Answer 63

\[G(x=3)=\log_4[2(3)+2]-2\]

Answer 64

but we are trying to find for x

Answer 65

ooh right.. my bad you are perfectly going well.. sorry to break your flow

Answer 66

lmao its okay i was kinda stuck anyways. do you add 2 to 3 to get \[5=\log _{4}(2x-2)\]

Answer 67

good..
now, how do you get rid of the "log (base 4)"?

Answer 68

do you do ln of 4

Answer 69

opposite of log is exponentiation and vice versa
so, you exponent the equation to base "4" since that is the base of your "log"
\[4^5=(2x+2)\]

Answer 70

okay so then do you -2 and then divide by 2

Answer 71

perfect

Answer 72

511?

Answer 73

yes!!

Answer 74

the point would be (511,3)

Answer 75

yep

Answer 76

so how would we find the zero

Answer 77

the zero of G

Answer 78

it means, for what value of "x" is G(x)=0?

Answer 79

so, set G(x)=0 and solve for "x" like part (c)

Answer 80

i got -31/32

Answer 81

nope..
steps?

Answer 82

i did 0=log base 4 (2x+2)-2
2=log base 4 (2x+2)
4^2=2x+2
7=x
opps did it wrong the first time. so 7 would be the answer

Answer 83

right.

Answer 84

haha thank you super much! is there chance you can help me with 2 more problems?

Answer 85

I have a good friend who is very nice.

Answer 86

post your question as a new one and I'll refer her there