# Solve inequation with nth root

I have the following inequation :

$\sqrt[n]{10} \geq \frac{10}{9}$

and I would want to know for which interval of n the inequation is right, but I have no idea how to solve it. I hope somebody could help me.

## 2 Answers

Take logarithms and get $n < \frac{\ln(10)}{\ln(10/9)} \approx 21.8543$. So if $n$ is an integer the inteval is $1 \le n \le 21$, and you can check $\sqrt{10} \approx 1.11588 > 10/9$ while $\sqrt{10} \approx 1.110336 < 10/9.$

• Thanks for your answer. It is exactly what I wanted. But just a question how you know $n < \frac{ln(10)}{ln(10/9)}$ ? – Lucas Willems Oct 23 '13 at 8:02
• Since $\ln(x)$ is a monotone increasing function you have (assuming $n>0$) $$\sqrt[n]{10} \ge \frac{10}{9} \iff 10^{\frac{1}{n}} \ge \frac{10}{9} \iff \frac{1}{n}\ln(10) \ge \ln(10/9) \iff \frac{\ln(10)} {\ln(10/9)} \le n$$ – gammatester Oct 23 '13 at 8:41
• Unfortunately the last $\le$ in my previous comment should also be a $\ge$, but I cannot edit any longer. – gammatester Oct 23 '13 at 11:40

Rewrite your left hand side term as 10^(1/n). Take the log of both side and get the inequality for "n"