trying to solve ln [ ln (ln x)]=1 ? .... I got to ln 0=1?
\[\Large \ln\left[\ln\left(\ln x\right)\right]\quad=\quad 1\]Are you familiar with the process of exponentiation? Or are you just jumping from log form to exponential form? Does this process make any sense to you?\[\LARGE e^{\ln\left[\ln\left(\ln x\right)\right]}\quad=\quad e^1\]
I really have not worked log form that much, so I'm fairly familiar but i'm not understanding the base e example? could you simplify? . thank you
The exponential and log function are inverses of one another. \[\LARGE \color{royalblue}{e}^{\color{royalblue}{\ln}\left[\ln\left(\ln x\right)\right]}\quad=\quad e^1\]So these two functions will essentially "undo" one another. Simplifying the problem to,\[\Large \ln(\ln x)\quad=\quad e^1\]
Is that a little too confusing? We can approach the problem a little differently if that doesn't make sense to you.
actually that does, but for some reason i'm wondering if ln (ln 1) =1wouldn't ln 1=0?
Yes ln 1 = 0 But we don't have ln 1 on the inside, we have ln x, don't we? :O
oh, ok. yes, sorry bout that.
This is going to seem like a weird process :) We're going to repeat what we did before to get rid of the outer log.\[\Large \ln(\ln x)\quad=\quad e\] Exponentiating each side gives us,\[\LARGE e^{\ln(\ln x)}\quad=\quad e^e\]
ok. I think i'm getting it, ln(lnx)=e . so how did it become e^1?
From our initial setup:\[\Large \color{#3366CF}{\ln\left[\ln\left(\ln x\right)\right]\quad=\quad 1}\]We want to rewrite each side as an exponent with a base of e,\[\LARGE e^{\color{#3366CF}{\ln\left[\ln\left(\ln x\right)\right]}}\quad=\quad e^{\color{#3366CF}{1}}\]
Then the left side simplified down,\[\Large \ln\left(\ln x\right)\quad=\quad e^{\color{#3366CF}{1}}\]
aaahhh.... : ) Ok. Crystal Clear, thank you very much. I appreciate it. I'm truly a genius. thanks.
lol :)
You understand where this process is going? You would have to "exponentiate" 3 times to get to your answer.
Join our real-time social learning platform and learn together with your friends!