Question

log_{1/3} (log_{4} (x^2-5)>0

Answer 1

@satellite73 u answer this yesterday can u explain the diff bet toays and yester

Answer 2

todays has one log
this one has two

Answer 3

so ..........

Answer 4

so we have to do it in two pieces

Answer 5

if u can plzz can u do it again..... if u r not intrested plzz inform me

Answer 6

can you find the question you asked last time? it should be in your profile

Answer 7

let me check it out and will inform u

Answer 8

we can do it again if you like
first of all don't forget that the domain of any logarithmic function is \((0,\infty)\) i.e. you cannot take the log of a non- positive number
so one thing we have to make sure of is that \(\log_4(x^2-5)>0\)

Answer 9

that has nothing to do with the inequality that you have to solve, that has to do with the domain of the function
we can solve that easily
\(\log_4(x^2-5)>0\iff x^2-5>1\iff x^2-6>0\iff x<-\sqrt{6}\) or \(x>\sqrt{6}\)

Answer 10

so we have that restriction before we even start to solve the inequality you are given, namely
\[\log_{1/3} (\log_{4} (x^2-5))>0\]

Answer 11

@satellite73 plz look my new ques http://openstudy.com/study?signup#/updates/4fbe5c0be4b0c25bf8fc84dd

Answer 12

now we also know that if
\(\log_{\frac{1}{3}}(x)>0\) we must have \(x<1\) so we also need to solve
\[\log_4(x^2-5)<1\]
\[x^2-5<4\]
\[x^2-9<0\]
\[-3<x<3\] put these together to get
\[\sqrt{6}<x<3\]