Observe that \(2^3 = 8\), \(2^5 = 32\), \(3^2 = 9\) and \(3^3 = 27\). From these facts, we can deduce that \(\log_{2}3\), the logarithm of \(3\) to base \(2\), is

between \(1\frac{1}{3}\) and \(1\frac{1}{2}\);

between \(1\frac{1}{2}\) and \(1\frac{2}{3}\);

between \(1\frac{2}{3}\) and \(2\);

between \(2\) and \(3\).

We are told that \[2^3 < 3^2 < 3^3 < 2^5.\]

As \(\log_{2}x\) is an increasing function (which means that \(\log_2x < \log_2y\) if and only if \(x < y\), or the gradient of the graph \(y = \log_2x\) is always positive), we can take logarithms to base \(2\) of these inequalities. This gives \[3 < 2\log_{2}3 < 3\log_{2}3 < 5.\]

From the first inequality we get \[\frac{3}{2} < \log_{2}3,\] and from the last inequality we get \[\log_{2}3 < \frac{5}{3}.\]

Combining these gives \[\frac{3}{2} < \log_{2}3 < \frac{5}{3},\] so the answer is (b).

An alternative approach might say, let \(\log_23 = k\), so \(3 = 2^k\). Now \(9 = 2^{2k} > 8\), so \(2k > 3\).

We also have \(27 = 2^{3k} < 32 = 2^5\), so \(3k < 5\).

We use here that \(y = 2^x\) is an increasing function.