Question

I need Help

Answer 1

@hartnn can u help me?

Answer 2

what have you tried?

Answer 3

I just can't start
I tried substituting every 2^x by

Answer 4

log(base2)2^x

Answer 5

let 2^x = X for simplicity.

y= X/(1+X)

got this?
can you isolate X?

Answer 6

what do u mean by isolating x

Answer 7

to find the inverse function of y = f(x)
we need to isolate x,
right?

something like x = ... (terms only containing y)

Answer 8

aha

Answer 9

I will try

Answer 10

x= 1-y

Answer 11

soorry
x=y-1

Answer 12

nope

Answer 13

did you get this?

et 2^x = X for simplicity.

y= X/(1+X)

Answer 14

yea

Answer 15

good,

y = X/(1+X)
fir starters,
multiply both sides by (1+X)

Answer 16

y(x+1) =x

Answer 17

now distribute
y(X+1) = ... ?

Answer 18

yx+y =x

Answer 19

then switching?

Answer 20

yes,

get the equation in the form
X = ... (only y terms)

Answer 21

x = y/ y+1

Answer 22

sure?

Answer 23

yea

Answer 24

I switched y and x then taking x as a factor

Answer 25

then i divided by y+1 in both sides

Answer 26

X = (1+X)y

X = y + Xy

X - Xy = y

X(1-y) = y

X = y/(1-y)

at which step did you make the error? :)

Answer 27

by switching y and x, you just replaced x by y and y by x?

Answer 28

yea

Answer 29

then you'll have to isolate y

and you should get
y = X/(1-X)

Answer 30

but don't do that, it'll create confusion

Answer 31

X = (1+X)y

X = y + Xy

X - Xy = y

X(1-y) = y

X = y/(1-y)

Did you get this?

Answer 32

yea

Answer 33

I see

Answer 34

now X = 2^x
remember?

Answer 35

2^x = y/(1-y)

take log on both sides!

Answer 36

how can I take the log on the side of y/1-y

Answer 37

right sides will simply be
\(\ln (\dfrac{y}{1-y})\)

Answer 38

what did u get for left?

Answer 39

x

Answer 40

but why ln on the right side
or did u mean log

Answer 41

i meant logarithm,
you could take any base,
ln has base e
taking log to the base 2 was correct :)

Answer 42

so you took log with the base 2 :)

finally you have
\(x = \log_2 (\dfrac{y}{1-y})\)

actually, thats it!

Answer 43

aha then the last step is switching

Answer 44

yessss!

Answer 45

Thanks buddy for helping me

Answer 46

most welcome ^_^
you could do part b) ?

Answer 47

sorry i got another question

Answer 48

ok

Answer 49

what is the domain

Answer 50

of this inverse we got

Answer 51

domain is all the values that x can take.

you're asking part (b), right?

Answer 52

yea

Answer 53

log of negative numbers is not defined.

so,
y/(1-y) must be positive.

makes sense?

Answer 54

yea

Answer 55

and 1 is not defined

Answer 56

2 cases:

1. both numerator and denominator must be positive
2. both numerator and denominator must be negative

wanna try on your own first?

Answer 57

yea

Answer 58

1. y> 0 and
1-y > 0
...

OR

2. y<0 and
1-y < 0
...

go ahead!

Answer 59

let me know when you get the domain :)

Answer 60

if you've already switched x and y,

1. x>0 and
1-x > 0
...

OR

2. x<0 and
1-x < 0
...

Answer 61

]0,1[

Answer 62

[0,1]

or
(0,1)
?

Answer 63

I wanna know something

Answer 64

note : ln 0 is not defined in real

Answer 65

(0.1)

Answer 66

yes, (0,1) :)

what did you wanna know?

Answer 67

1. x>0 and
1-x > 0
...

OR

2. x<0 and
1-x < 0


why did u wrote number 2?
and i think number 2 doesn't make sense
how to solve it

Answer 68

-3/-4 is positive, right??

Answer 69

right

Answer 70

a/b is positive if

a and b both are positive
OR
a and b both are negative

Answer 71

thats why we have to consider 2. also.

x<0 and 1-x<0

Answer 72

aha
but 2 is wrong in this example or I mean that it didn't make sense, right?

Answer 73

now
1-x < 0
means
x>1

BUT,
x<0 "and" x>1 is simultaneously NOT possible.

hence we do not get any solutions from 2.

Answer 74

see if that makes sense ^^

Answer 75

yea

Answer 76

good :)

Answer 77

I see thanks bro

Answer 78

welcome ^_^

Answer 79

Can u help me in something

Answer 80

sure

Answer 81

I intended to take SAT II Math

Answer 82

Okay

Answer 83

So I am Studying from Pre-calculus James Stewart's book
is it good ?

Answer 84

I never took SAT... so I can't comment on the book, sorry...

Answer 85

thanks bro ^_^