Logarithm in inequality. Prove that $\log{n}\le\frac{n}{10}$ where $n\ge10$  and $\log{n}>\frac{n}{10}$ where $n<10$.    $n$ is a positive integer not equal to $1$.
I have tried this question by graphs. Is there any another method to approach this question?
 A: I will start with $1 < n < 10$
Since $n$ is an integer, we can simply test all 9 values, and see that it holds. 
for 2: $0.301... > 0.2$
for 3: $0.477... > 0.3$
for 4: $0.602... > 0.4$
for 5: $0.699... > 0.5$
for 6: $0.778... > 0.6$
for 7: $0.845... > 0.7$
for 8: $0.903... > 0.8$
for 9: $0.954... > 0.9$
Let's check $n=10$
$\log 10 = 1$
$\frac{10}{10} = 1$
They're equal, let's check for $n > 10$
Since we know that they're equal at $n=10$, we can see that $\log n < \frac{n}{10}$ if we show that for any $n > 10$, $\log (n+1) - \log n < \frac{n+1}{10}-\frac{n}{10}$
Let's start to see how much $\frac{n+1}{10}-\frac{n}{10}$ increases with:
$$\frac{n+1}{10}-\frac{n}{10}$$
$$\frac{n+1-n}{10}$$
$$\frac{1}{10}$$
So it increases with $\frac{1}{10}$ when you move 1 in the graph.
If we see how much $\log (n+1) - \log n$ increases with we can do this:
$$\log (n+1) - \log n$$
$$\log \frac{n+1}{n}$$
$$\log\left(\frac{1}{n} + 1\right)$$
Since $\frac{1}{n} + 1$ is decreasing as $n$ increases, $\log\left(\frac{1}{n} + 1\right)$ decreases as $n$ increases.
This means that if $\log\left(\frac{1}{n} + 1\right) < \frac{1}{10}$ at $n = 10$, then $\log\left(\frac{1}{n} + 1\right) < \frac{1}{10}$ holds for all $n > 10$
$$\log\left(\frac{1}{n} + 1\right) = 0.0414... < 0.1$$
A: You have $\log_{10}n = \frac n {10}$ for $n=10$.  $\;\log_{10}n$ grows like $1/(n \ln(10))$, whereas $n/10$ grows with constant $1/10$. So fo $n\ge 10$ your right side grows faster...
