Question

solveeeeee

Answer 1

\[\log_2{(\frac{x-5}{2x-3})^2}<0\]

Answer 2

I'd recommend using rule
\[\huge \log_ab=c \leftarrow \rightarrow b=a^c\]

Answer 3

but what is c here

Answer 4

0

Answer 5

\[\huge \log_ab=c \leftarrow \rightarrow b=a^c \\ \\ \huge (\frac{x-5}{2x-3})^2<2^0\]

Answer 6

yes.. i got the same

Answer 7

now, rhs will become 1

Answer 8

No you don't need it'll be just 1

Answer 9

okay

Answer 10

The sign should be x<-2

Answer 11

Why not expand it?

Answer 12

okay

Answer 13

expand
how

Answer 14

\[\huge (\frac{x-5}{2x-3})^2<2^0 \\ \\ \huge \frac{(x-5)^2}{(2x-3)^2}<1 \\ \\ \huge x^2-10x+25<4x^2-12x+9\]

Answer 15

okay

Answer 16

you'll have to consider this actually :

-1 < (x-5)/(2x-3) < 1

Answer 17

one inequality at a time, 2 in total
and take intersection of both results

Answer 18

Its ok we'll have restrictions later

Answer 19

hmm

Answer 20

so what should i do
i mean should i follow @shubhamsrg or @.Sam. method

Answer 21

both are the same thing, @.Sam. is getting to it, he might guide you step to step :)

Answer 22

thank u

Answer 23

@.sam i got
3x^2-2x-16<0

Answer 24

Factor it

Answer 25

Yes then you got
x=-2 and x=8/3
Using

We have
x<-2 and x>8/3

Answer 26

okay

Answer 27

thank u @.Sam.

Answer 28

But from
\[\log_2{(\frac{x-5}{2x-3})^2}<0\]
x values can only be less than zero, so the only answer is x<-2

Answer 29

okay

Answer 30

\[\log_2{(\frac{x-5}{2x-3})^2}\color{red}{<0}\]

Answer 31

i got it