How to solve this inequality problem (regarding polynomal and log) I need to find what range of natural numbers satisfy this in-equality using calculus tools:
$ n < 8lg(n)$
Well, I set a function 
$f(x) = 8lg(x) - x\\$
Its log base 2.
And tried to use the derivative to find where $f'(x)>0$ but i dont see how it helps me.
Thank you.
 A: You can try to use the Lagrange Multiplier method with an slack variable, I recommend read this material: http://users.wpi.edu/~pwdavis/Courses/MA1024B10/1024_Lagrange_multipliers.pdf
A: I have a solution in the mind:
Assume you know the solution of $f(x)=0$ that is a positive $c>0$ such that $8\log_2(c)=c$. In fact,the continuous functions $2^x$ and $x^8$ on $(0^+, +\infty)$ intersect each other. By guessing or using a software $c$ is a number near to $1$ from the right. As you pointed, $f'(x)=\frac{8}{\ln(2)x}-1$ and decreasing on above interval. Now let $0^+<x<c$ so we deduce that $$0=f(c)<f(x)=8\log_2(x)-x.$$
So I think your answer would be empty.
A: If $x = 8\lg(x)$,
then,
thanks to Lambert the fearless function
and Wolfy,
$x_0 =e^{-W(-\log(2)/8}
≈1.09999703023760940090...
$
and
$x_1 =e^{-W_{-1}(-\log(2)/8}
≈43.559260436881656414...\\
$.
$8\lg(x) > x$
for
$x_0 < x < x_1$
and
$8\lg(x) < x$
for
$0 < x < x_0$
and
$x > x_1$.
A: Look at the derivative, it tells you where difference is increasing or decreasing. Pick some "easy" points in the respective ranges to bracket any potential zeros, see if you can narrow the ranges further.
