Without using a calculator or computer system, determine the following (you will need to know some properties of logs to answer this question):
(a) $⌈\log_{10} (101)⌉$
(b) $⌊\log_2 (63)⌋$
(c) $⌊\log_n (n^x)⌋$ where $n$ is a natural number and $n− 1 < x < n$.
(a) We first recognise
$$ 100 = 10^2 < 10^3 = 1000 $$
$$ \log_{10} 100 = 2 < 3 = \log_{10} 1000 $$
We used the fact that a logarithm is monotonically increasing. This also means
$$ \log_{10} 100 = 2 < \textcolor{red}{ \log_{10} 101} < 3 = \log_{10} 1000 $$
And so
$$ \lceil \log_{10} 101 \rceil = 3 $$
because 3 is the smallest integer larger than $ \log_{10} 101$.
(b) We first recognise
$$ 32 = 2^5 < 2^6 = 64$$
$$ \log_2 32 = 5 < 6 = \log_2 64$$
Again, we used the fact that a logarithm is monodically increasing. This also means
$$ \log_2 32 = 5 < \textcolor{red}{ \log_2 63} < 6 = \log_2 64$$
And so
$$ \lfloor \log_2 63 \rfloor = 5 $$
because 5 is the largest integer smaller than $\log_2 63$.
(c) We are given $n− 1 < x < n$, where $n$ is a natural number, and so
$$ n^{n-1} < n^x < n^n $$
Since logarithms are monotonically increasing,
$$ \log_n n^{n-1} < \log_n n^x < \log_n n^n $$
Simplifying
$$ n-1 < \log_n n^x < n $$
And so
$$ \lfloor \log_n n^x \rfloor = n-1 $$
because $n-1$ is the largest integer smaller than $\log_n n^x$.