Question

Sollution set of

Answer 1

\[\huge \log_x(2+x) \le \log_x(6-x)\]

Answer 2

Can't we cancel \(\log_x\) both the sides or by taking anti-logarithm?

Answer 3

As a start : change to natural log base and cancel some stuff maybe

Answer 4

you don't know base haha

Answer 5

we cannot cancel something without knowing its sign

Answer 6

or , i was thinking of taking two conditions
one which will change sign
one which will not
and take intersection

Answer 7

Its sign?? What is "its' here??

Answer 8

you can't just cancel the logs ,

Answer 9

i wanted to make sure

Answer 10

that is not giving the anser

Answer 11

I am asking Ganesh.. :P

Answer 12

considering two cases sounds like a nice plan

Answer 13

the plan failed

Answer 14

something = base, water :)

Answer 15

i did it , i am getting
x ge 0
x le 0

Answer 16

from the input inequality
x>0
x+2>0
6-x>0

yes ?

Answer 17

Between -2 and 6..

Answer 18

that bounds the possible solutions to
\[\large 0\lt x\lt 6\]

Answer 19

we still need to find out the actual solutions

Answer 20

yeah answer is
(1.2]

Answer 21

Yes you are right..

Answer 22

when is logx negative ?

Answer 23

Between 0 and 1??

Answer 24

yeah

Answer 25

When x is in decimal form..

Answer 26

\[\large \log_x(2+x) \le \log_x(6-x) \]

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

Answer 27

so you don't have to flip inequality when you multiply lnx for x>1, yes ?

Answer 28

case 1 :
0<x<1

case 2:
1<x<6

Answer 29

yes

Answer 30

Leave it , i will ask my prof.

Answer 31

aren't we done ?

Answer 32

case 1 :
0<x<1
\[\large 2+x \ge 6-x \implies x \ge 2\]

but we assumed that 0<x<1
so no solutions for case 1

Answer 33

case 2 :
1<x<6
\[\large 2+x \le 6-x \implies x \le 2\]


\(\large (x\le 2 ) \wedge ( 1\lt x\lt 6) \implies 1\lt x\le 2 \)

Answer 34

yup , we were done