Question

Anyone?

Answer 1

simplify using log properties ?

Answer 2

\[\large a \ln b = \ln b^a\]

\[\large \ln a - \ln b = \ln\left(\dfrac{a}{b}\right)\]

Answer 3

i need a set of solutions for this inequality
i know those..but i get stuck at some point

Answer 4

where exactly you got stuck

Answer 5

so i did this:
2ln(1-x)-ln(2x+6)<=0
2ln(1-x)/ln(2x+6)<=0
2ln(1-x)<= 0 and 2ln(2x+6)>=0
and idk what to do with this anymore xD

Answer 6

lets work it step by step

Answer 7

lets :D

Answer 8

\[\large 2 \ln(1-x) - \ln (2x+6) \le 0\]

Answer 9

step 1 : for the first term, use the log property \(\large a \ln b = \ln b^a\)

\[\large \ln(1-x)^2 - \ln (2x+6) \le 0\]

Answer 10

okay..

Answer 11

next use the second log property

\[\large \ln \left(\dfrac{(1-x)^2}{ (2x+6)} \right)\le 0\]

Answer 12

still wid me ? :)

Answer 13

yup

Answer 14

next change it to exponent form using below :

\(\ln a = b \) means \(a = e^b\)

Answer 15

\[\large \ln \left(\dfrac{(1-x)^2}{ (2x+6)} \right)\le 0\]

\[\large \dfrac{(1-x)^2}{ (2x+6)} \le e^0\]

\[\large \dfrac{(1-x)^2}{ (2x+6)} \le 1\]

\[\large (1+x)^2 \le 2x+6\]

Answer 16

see if you can solve that inequality ^

Answer 17

\[\large (1\color{Red}{-}x)^2 \le 2x+6 ~~^*\]

Answer 18

so i have x^2-4x-5<=0 and then i just do it like a normal equation right? :D

Answer 19

yes factor it first

Answer 20

english is my 3rd language so i'm not quite sure what that means >_<

Answer 21

can you factor the quadratic ?

Answer 22

x^2-4x-5<=0

(x )(x )<=0

Answer 23

right right i do xD
we just do that differently

Answer 24

x1/2=... thing :D

Answer 25

ohkay, what do u get after u factor it ? :)

Answer 26

5 and -1

Answer 27

since the leading coefficient is positive, the graph between these zeroes stays negative, so the solutions will be :

-1 <= x <= 5

Answer 28

however, from the first term of given given equation, x must be less than 1 (why ? )

Answer 29

but it says here that the correct answer is [-1,1) :/

Answer 30

so the final solutions will be :

-1 <= x < 1

Answer 31

(1−x)2/(2x+6)≤1
because of this or..

Answer 32

\[\large 2 \color{red}{\ln(1-x) }- \ln (2x+6) \le 0\]

Answer 33

that part ^

Answer 34

since log cannot suck in negative values or 0 :

1-x > 0

x < 1

Answer 35

got it! thank you :)