# Show that for each number $n > 0$, $nx^{1/n} < \ln(x)$ for $x > 1$

Show that for each number $n > 0$ $$nx^{1/n} < \ln(x)$$ for $x > 1$.

Update as pointed below, original problem set has a misprint, and the inequity should be

$$nx^{1/n} > \ln(x)$$

I'm working on a problem set for MIT Open Courses, Single Variable Calculus, Unit 2 - Applications of Diﬀerentiation, Mean Value Theorem section, problem 2G-7 C. (included for future google searches).

• Can you put a backslash in front of $\ln$ in your MathJax – Don Thousand Aug 16 '18 at 20:18
• This isn't true, since $n\cdot 1^{1/n} = n > 0 \log(1)$ for $n>0$. Are you sure you don't mean $n x^{1/n} > \log x$? – cdipaolo Aug 16 '18 at 20:18
• Nope, not sure at all, since I've found and confirmed errors in OCW problems sets on more than one occasion. Hence Im asking here after googling gave no results. – Ivan Koshelev Aug 16 '18 at 20:21
• @cdipaolo, the OP has accurately presented the problem as it appears in the link. The mistake is in the problem set. As you indicate, the inequality should point in the other direction. – Barry Cipra Aug 16 '18 at 20:25

If you divide by $n$, the LHS is $x^{1/n}$, and the the RHS is $\ln(x^{1/n})$. Now $a>\ln a$ iff $1<a$, so the inequality holds in the reverse direction.
I'm assuming you mean to prove $n x^{1/n} > \log x$ when $n > 0$ and $x>1$ (the other direction is false). To see this, observe that $n\cdot 1^{1/n} = n > 0 = \log 1$. Moreover, when $x\geq 1$, $$\frac{d}{dx}nx^{1/n} = x^{\tfrac{1-n}{n}} \geq \frac{1}{x} = \frac{d}{dx}\log x.$$ Together, this means that $$n x^{1/n} = n + \int_1^x \frac{d}{dt}n t^{1/n}\,dt > 0 + \int_1^x \frac{d}{dt}\log t\,dt = \log(x)$$ as desired.