Question

can some1 help me to find the domain of the following function f(x)=log(logx)

Answer 1

show the work please

Answer 2

Yeah I can
Now \(logx > 0\)

Answer 3

If \(logx > 0 \)
\(\text{Then}\)
\(x > e^0\)
\[x>1\]

Answer 4

Don't get it! how do you get at x > e0?

Answer 5

\(log_ex = 2\)\[\text{Then}\]\[x = e^2\]

Answer 6

He raises both sides to e. But since it's log which is base 10 it should be:
\[logx>0\]
\[10^{logx}>10^0\]
Log cancels with 10:
\[x>1\]
Since x^0 always = 1.

Answer 7

The domain is the allowed x values.
log(x) may not have x negative. (looking at the outermost log)
Thus log(x)>0. Thus x>1

Answer 8

Blah, dumb line breaks.

Answer 9

\[Let\ f(a) = log(a)\]\[\implies Domain_f : a > 0\]
\[\large \therefore f(log(x)) = log(log(x))\]\[\large\implies Domain_{f(log(x))}: log(x) > 0\]\[\large \implies x > 10^0 \implies x > 1\]

Answer 10

wow

Answer 11

Why I cannot do the following?

logx>0 puting this expression into a exponential one
\[10^{0} > x\]

so x<1

I know x>1 is correct. But tell me how to figure out the other way is not correct?

Answer 12

You flipped the sign. For inequalities the only time the sign flips is if you divide by negative numbers.

Answer 13

So when you moved the x to the right the sign should have followed it as such:
\[10^0<x\]

Answer 14

ahhh got it ;)