Question

if "1 > x"

is x > 1
or
is x < 1

may sound too easy, but this is troubleshooting me a lot.

Answer 1

if x is less than 1, then x is less than 1 regardless of where you put it

Answer 2

the symbol is simply defining the relation; it isnt set in stone; if x moves, move the symbol

Answer 3

think of it as xs little flashlight :) or whatever number/object is using it

Answer 4

Cannot remember, but some1 told me that in the expression "1>x"

if you change the x to the left hand side and the 1 to the left hand side simultaneously, you have to change the direction of the inequality.

that is: x<1
................................................

Today I had to find the domain of the following function: f(x) = Log(Log(x))

I made Log(x) > 0
then I put in into an exponential expression, that is:

10^0 > x
1>x

and I flipped the direction of the sign, so I end up with

x<0

Answer 5

i dont think you can create a log(log(x)) function tho

Answer 6

yeah but you want
\[\log(x)>0\] right?

Answer 7

exactly

Answer 8

yeah that caveat

Answer 9

no that is wrong

Answer 10

there is some trick teachers will play saying find this or that and give a function that makes no sense

Answer 11

on no
\[\ln(\ln(x))\] if a perfectly good function

Answer 12

I need to find the domain of f(x) = Log(Log(x))

Answer 13

I know the correct answer is x>0

Answer 14

an nice example of a function whose limit as x goes to infinity is infinity, but that grows very very VERY slowly

Answer 15

a no, that is not the right answer

Answer 16

sorry the correct answer is x>1

Answer 17

you cannot take the log of a negative number.

Answer 18

right that is the correct answer

Answer 19

you have to make sure that
\[\log(x)>0\] meaning
\[x>1\] done

Answer 20

for clarity tho

http://www.wolframalpha.com/input/?i=domain+log%28log%28x%29%29

Answer 21

@amistre this is
\[\ln(\ln(x))\] and it is graphing real and imaginary parts.

Answer 22

this is not the "real number" graph.

Answer 23

wolfram did it, not me :) i cant get it to do what i want it to :)

Answer 24

http://www.wolframalpha.com/input/?i=y%3Dlog%28log%28x%29%29++real

Answer 25

the plot is this, right?

Answer 26

i specified "real" and it gave the domain too!

Answer 27

probably. it looks like a log

Answer 28

the picture amistre sent has it, just ignore everything to the left of 1, where it is graphing real and complex parts

Answer 29

log(x) > 0; when x > 10^0, which is when x>1

Answer 30

You simplified the log by setting one base in in both sides.

Answer 31

yeah, i think

Answer 32

log(1) = 0 ; so everything past it

Answer 33

I know that way. I want to carry out the the same problem by changing it to the exponential form

Answer 34

\[log(log(x))=y\]
\[10^{log(log(x))}=10^y\]
\[log(x)=10^y\]
\[10^{log(x)}=10^{10^y}\]
\[x =10^{10^y}\]
something like that?

Answer 35

remember that I need to find the domain of the function. THat is , I have to limit the inner log to Log(x) > 0

so I get 10^0 > x. then I get 1 > x. That means that if I want to get to the correct answer I must change x to the left size and the 1 to the right side and no changing the direction of the signg.

Answer 36

the domain of the function is the range if its inverse :)

the range of its inverse is (1,inf)

Answer 37

\[\lim_{y->-\infty} 10^{10^y}=1\]

Answer 38

\[log(x)>0\]
\[10^{log(x)}>10^0\]
\[x>1\]
as long as x is greater than 1 your good

Answer 39

your either a little dyslexic in your process; or your making a simple mistake

Answer 40

I think the second one haha

Answer 41

\[log(x)>0\]
does not morph into\[10^0>x\]

Answer 42

that means 1 > x or x > 1, right?

Answer 43

no; it means that x >1 , which is the same as 1 < x

Answer 44

1 < x < inf\[\]
inf < x < 1

Answer 45

Im sorry if this is getting boring, but 10^0 > x is the same as 1 > x. Right?

Answer 46

missed that one lol; ignore the typo

Answer 47

where are you getting the 10^0 > x from? your inventing that from some nonmath standard and trying to fit it into the problem.

Answer 48

No. I want to find the domain of the function f(x) = Log(Log(x)). Logs can only be possitive, so I made Log(x)>0.

Answer 49

good; log(x) > 0 , the next step is?

Answer 50

put it into an exponetial form. that is 10 ^0 > x, right?

Answer 51

exponentiate!

Answer 52

but you have the inequality backwards!

Answer 53

\[log(x)>0\]
invent some new form of non math procedure and come up with
\[10^0 > x\]
what steps are you using to invent this abomination?

Answer 54

\[\log(x)>0\]
\[10^{\log(x)}>10^0\]
\[x>1\]

Answer 55

i am using "exponentiate"

Answer 56

use the rules that math provides, and you come to a consistent answer everytime; invent your own rules and you will end up who knows where :)

Answer 57

the function
\[g(x)=10^x\] s a strictly increasing function so if
\[a>b\] if follow that
\[10^a>10^b\]

Answer 58

That is what is throublering me.

In the equation log(x) = 0, you can do the following: 10^0 = x; can I do the same with inequalities?

Answer 59

you can as long as you see that you flipped at all around; along with the equals sign

Answer 60

that sign is as mobile as the others ... it just doesnt make that much of an impact visually

Answer 61

log(x) = 0\[\]
log(x) > 0\[\]

0 = log(x) ; did you see the equal sign flip around too?\[\]
0 < log(x)\[\]

10^0 = x\[\]
10^0 < x\[\]

Answer 62

make it a general rule; when your flipping things aroung; flip your signs\[\]
when you multiply or divide by a "-"; flip your signs .... all of them

Answer 63

dont try to remember separate rules for separate things when 1 rule does it all

Answer 64

does that make sense?

Answer 65

yeah finally :) Normally I understand maths quicky, but this time ups ... Thank you very much amistre64 and Satellite

Answer 66

would you think this is good form?\[\]
x < 3\[\]
x = 3\[\]
3 = x\[\]
3 < x\[\]
??

Answer 67

No. I think 3 > x. You have to flip the sign, right?

Answer 68

you flipped everything around; including the sign ... no matter what the sign looks like, =, >, < , any of them

Answer 69

so my answer was right?

Answer 70

yes, your answer was right :)\[\]

what property do we have that says:\[\]

x = 3 therefore 3=x?\[\]

\[x=3\]
\[x-3=3-3\]
\[x-x-3=0-x\]
\[-3=-x\]
\[-3(-1)=-x(-1)\text{ ; right here we have to flip the sign right?}\]
\[3=x\]

Answer 71

try it with your > and < signs and see if you need to :)

Answer 72

how can I pass to the other line? I give enter but everything is shown in one single line

Answer 73

force a carriage return by putting in an empty equation:\[\]

\ then [ then \ then ] \[\]
at the end of where you want it to break

Answer 74

h= 10 +t \ then [ then \ then ]

Answer 75

???

Answer 76

take out the "thens" of course ..

Answer 77

if i try to just type it in to show you it doest\[\]show up ...

Answer 78

h= 10 +t \ [ \ ]

Answer 79

so i have to break it apart so it doesnt get feed to the parser as an equation; now remove all the spaces

Answer 80

ok I think this problem will be solved tomorrow or the day after. So im gonna use mathematica worlfram for the moment. Wait and Im gonna try x=3 with the '>' ;)

Answer 81

you do that with the >; and it should convince you that your "shorcut" is really just plowing over the real math that is usually done behind the scenes :)

Answer 82

is that right?