Find least positive integer $n$ which satisfies the condition Find least positive integer $n$ which satisfies the condition 


$| {0.5}^{\frac{1}{n}}-1|&lt;10^{-3} $
$0.9^{n}&lt;10^{-3}$


What I have tried is I have split the first inequality  took log on both side .. but I came to know it is getting complicated
for the second one we get by taking the logarithm $$n\ln\left(\frac{9}{10}\right)&lt;\ln(10^{-3}$$ therefore $$n&gt;\frac{10^{-3}}{\ln(0.9)}$$
we get $$n\geq 66$$
for the second one we get
$$n&gt;\frac{\ln(0.5)}{\ln(1-10^{-3})}$$ therefore $$n\geq 693$$
\begin{array}{c}
   | 0.5^{\frac 1n}-1| &lt; \frac{1}{1000}\\
   1-\frac{1}{1000} &lt; 0.5^{\frac 1n} &lt; 1+\frac{1}{1000} \\
   \frac{999}{1000} &lt; 0.5^{\frac 1n} &lt; \frac{1001}{1000} \\
   \log 999 - 3 &lt; -\frac 1n \log \dfrac 12 &lt; \log 1001 - 3 \\
   \log 999 - 3 &lt; \frac 1n \log 2 &lt; \log 1001 - 3 \\
   \dfrac{\log 999 - 3}{\log 2} &lt; \frac 1n &lt; \dfrac{\log 1001 - 3}{\log 2} \\
\end{array}

$\dfrac{\log 999 - 3}{\log 2} &lt; 0 &lt; \frac 1n$ is always true for positive integer values of $n$.

\begin{align}
   \frac 1n &lt; \dfrac{\log 1001 - 3}{\log 2} \\
   n &gt; \dfrac{\log 2}{\log 1001 - 3} \\
   n &gt; 693.49\dots \\
\end{align}

The least integer value of $n$ is $n = 694$