Question

solve for x:

Answer 1

x...

Answer 2

\[\log_{2}(x+1)/(x-1) > 0 \]

Answer 3

uhmm let me get this straigh...yes or no...
\[\huge \log_2 (\frac{x+1}{x-1}) > 0?\]

Answer 4

numerator - log{base2} (x+1)
denominator - x-1

Answer 5

\[\log\frac{a}{b}=\log a-\log b\]

Answer 6

@UnkleRhaukus it is log only in numerator.

Answer 7

\[\huge \frac{\log_2 (x+1)}{x-1} > 0?\]

Answer 8

yes

Answer 9

oh

Answer 10

this seems harder than i thought :/

Answer 11

cross multiply...what do you get?

Answer 12

you can't cross multiply a variable in an inequality.

Answer 13

i know...but i cant see anything easier haha lol

Answer 14

@shubhamsrg @Wired Help!!

Answer 15

see for a fraction to be greater than zero both num & shud be be of the same sign

Answer 16

x>1 or x<0

Answer 17

see..we have to find those values of x for which LHS >0
if -1<x<0
num = -ve
denom = -ve
result = +ve
note that -1 and 0 are not included
for 0<x<1
num = +ve
denom =-ve
result -ve
hence (0,1) is not a part
clearly, for x>1, LHS is +ve
hence ans should be (-1,0)U(1,infinity)
NOTE for x<-1,,log becomes undefined

Answer 18

(-1,0)U(1,infinity)

Answer 19

@shubhamsrg how have you decided "-1<x<0" and "0<x<1" ?

Answer 20

the log is defined only for numbers> 0, so (x + 1)>0, and x> -1

Answer 21

well only these points/numbers seem to be the one where there is a chance of break/discontinuity etc..