Question

use the one to one property tp solve log_2(x-3)-log_2(2x+1)=-log_2(4).. the answer in the book is 13/2.. but i dont know how to get there so please show work!! thank you!!

Answer 1

@phi .. you helped me last time and was very helpful.. any ideas on thsi one?

Answer 2

on the right side log_2(4) means what number x is it so that 2^x = 4
any idea ?

Answer 3

2

Answer 4

\[ \log_2(x-3)-\log_2(2x+1)=-\log_2(4) \\
\log_2(x-3)-\log_2(2x+1)= -2
\]

Answer 5

On the left side, use log(a) - log(b) = log(a/b) to combine the two logs

Answer 6

okay ya i did that!

Answer 7

i got... log_2((x-3)/(2x+1))=-2 .. can i just cross out that log_2?

Answer 8

@phi

Answer 9

to get rid of a log, you make it the exponent of the base. in other words, do this
on both sides:
\[ 2^{\log_2(stuff)} = 2^{-2} \]

by definition, the left side turns into
\[ stuff = 2^{-2} \]

Answer 10

wait sorry im really confused at that point

Answer 11

if log_2((x-3)/(2x+1))=-2 then
(x-3)/(2x+1)=2^-2
solve for x

Answer 12

so on the right itd be .25?

Answer 13

yes, this part is confusing.

on the right side you get ¼ (or 0.25) but for the time being I would use ¼

the left side is confusing, but the idea is
if you have log_2(A)
you can "undo" the log_2 by making log_2(A) an exponent of 2
\[ 2^{\log_2(A)} \]
which *by definition* is A

Answer 14

so we make each side the exponent of 2 (to keep things equal)
the right side turns into ¼
the left side turns into (x-3)/(2x+1)

Answer 15

ok i got thattt now just solve?

Answer 16

wouldn't i just multiply both sides by (2x+1) next? so itd cancel out on the left and then distrbute on the right?

Answer 17

yes, sounds like a good plan
and multiply both sides by 4 to get rid of the fraction

Answer 18

what one should i do first?

Answer 19

order does not matter.
you can think of it as cross multiplying (if you like)

Answer 20

\[ \frac{(x-3)}{(2x+1)}=\frac{1}{4} \]

Answer 21

so on the right id just have 1.. then on the left id have,, (4x-12)/ 8x+8)?

Answer 22

ohh wait so i can just corss multiply that equation up top?

Answer 23

no, you multiply both sides by 4. the 4 is up top.

Answer 24

**no to (4x-12)/ 8x+8)?
yes to cross multiply.

Answer 25

so.... (4x-12)/(2x+1)

Answer 26

(4x-12)/(2x+1)=1
always write the whole equation

Answer 27

now multiply both sides by (2x+1)

Answer 28

so 8x^2-12=2x+1... right.. or do i foil

Answer 29

notice that (2x+1) times the left side will cancel out the (2x+1) on the bottom

Answer 30

you could multiply (2x+1)(4x-12) and use FOIL (which gives a different answer than what you got), but that is not what we want.

\[\cancel{ (2x+1)} \cdot \frac{4x-12}{\cancel{2x+1}} = (2x+1) \cdot 1 \]

Answer 31

thank youu!!!!!!1!

Answer 32

how about find the x intercepts of the function: f(x)=e^(x+4)-e... the answers in the book is (-3,0).. any idea how to perform it? @phi

Answer 33

x intercepts are where y ( i.e. f(x) ) is 0
so we have to solve
e^(x+4) - e^1 = 0
we could factor out e^1 from each term
remember \( e^{a+b}= e^a \cdot e^b \)
lets write x+4 as (x+3) + 1 so we have
\[ e^{(x+3)+1} - e^1 =0\\
e^{(x+3)}\cdot e^1-e^1 = 0
\]

Answer 34

ok got that!

Answer 35

what would i do nextrt

Answer 36

@phi

Answer 37

so then just (x+3)=0 and it would be -3?

Answer 38

starting with
\[ e^{(x+3)}\cdot e^1-e^1 = 0 \]
what do you get after factoring out e^1 ?

Answer 39

ummm e^(x+3)=0? wwouldnt both of the e^1 factor out?

Answer 40

what would you get if you factor a out of
(ba - a) = 0

Answer 41

a(b-1)=0

Answer 42

now notice a could be e^1 and b could be e^(x+3)

Answer 43

it is the same problem, with different symbols.

Answer 44

soo.. e^1(e^(x+3)-1)=0

Answer 45

yes. now divide both sides by e^1 (or multiply both sides by e^(-1) )

Answer 46

okk

Answer 47

or in
a(b-1)= 0
divide both sides by a

Answer 48

so.. (e^(x+3)-1)=e^1

Answer 49

0/e^1 is not e^1

Answer 50

0

Answer 51

\[ \frac{1}{\cancel{e^1}}\cdot \cancel{e^1}(e^{(x+3)}-1)=\frac{0}{e^1} \]

Answer 52

you now have
\[ e^{x+3}-1=0 \]
add 1 to both sides

Answer 53

ok

Answer 54

we have so far ?

Answer 55

e^(x+3)=1

Answer 56

@phi

Answer 57

any idea how to "bring down" the (x+3) ? I would take the natural log of both sides.

Answer 58

yaa i was gonna say that.. but then what happens to the ln1 on the right? how to i cancel the ln out

Answer 59

you don't cancel out ln 1
however, we can figure out what number that is. what x is it , so that e^x = 1 ?

Answer 60

e^0 =1

Answer 61

sweet thanks i got it!!

Answer 62

and ln(e^0) = ln 1
or
0 * ln(e) = ln(1)
0 = ln(1)