"Find x 3 such that ln(x)

Question

"Find x > 3 such that 

ln(x) < x^(0.1)"


How do you solve this type of problem? I plugged it into my TI-89 solver and didn't get anything. As a hint, the number is unbelievably huge.

anonymous · Answer

my teacher gave us this problem in class today and no one knew what to do can someone help?

anonymous · Answer

seems hard can someone hekp me http://openstudy.com/study#/updates/4fc65a22e4b022f1e12dd2b9

anonymous · Answer

$$
\log x<x^{0.1}\
\log_x\log x<0.1\
\frac{\log\log x}{\log x}<0.1
$$There's a start, anyway.

anonymous · Answer

ok I have it but at least I know there is an answer now have to find out why
 x=8.8E+15 giving 
ln(x) = 36.71
x^(0.1) = 39.3

anonymous · Answer

@nbouscal the question was natural log not log so its to the base e not 10?

anonymous · Answer

When I say $\log$ I mean the natural logarithm. If I mean $\log_{10}$ I'd say it :P

anonymous · Answer

ok then say ln as that's what its called :P otherwise there is confusion!

anonymous · Answer

That's only what it's called in fields where using base 10 logarithms is useful in some way. In pure mathematics, 10 is just a number, it's not special. The default assumption should be the natural logarithm.

anonymous · Answer

Ah to change ln to log  you divide by the conversion factor 2.303, might be good as you can't have ln with a base of x. the conversion to base 10 could be a start?

anonymous · Answer

Why would you convert to base 10? There's no mathematical reason to work in base 10. I already got rid of the logarithm with base x, when I simplified it to $$
\frac{\log\log x}{\log x}<0.1
$$

anonymous · Answer

ok, let me get what you did

anonymous · Answer

ok I see what you did now what

anonymous · Answer

Well, now we can subtitute $t=\log x$, so we have $\dfrac{\log t}{t}<0.1$

anonymous · Answer

Solving $\dfrac{\log t}{t}<0.1$, we learn that $t>35.7715$, approximately. So we know that $\log x >35.7715$. That means that $x>e^{35.7715}$, or approximately $x>3.43\times10^{15}$.

anonymous · Answer

That looks good, and its in the same ball park as my estimate. Nice job :D

anonymous · Answer

The only question is how we solve $\log t<0.1t$. I used WolframAlpha, which in turn used the Lambert W-function, which is a relatively advanced technique. I'm sure we could have got there using a kind of guess and check, I'm just not sure whether that was the method expected by the teacher in this case.

anonymous · Answer

Sorry back now, so how did you get t>35.7715 ? was that using WolframAlpha?

anonymous · Answer

ok, got there using excel and seen it on Wolfram.hmmmm