The expected value of a discrete random variable is given by

\mathop{\mathrm{E}}(X) = \sum x \, \mathop{\mathrm{P}}(X = x) \,\,\,\,\,\left\lbrace\text{SL 4.7}\right\rbrace

Substitute the given probabilities:

\mathop{\mathrm{E}}(X) = 1 \cdot \ln\left(\frac{5}{4}\right) + 2 \cdot \ln\left(\frac{6}{5}\right) + 3 \cdot \ln\left(\frac{4}{3}\right) + 4 \cdot 2\ln k

Apply logarithm properties (a\ln x = \ln x^a, \ln a + \ln b = \ln(ab), \ln a - \ln b = \ln(a/b)) \left\lbrace\text{SL 1.7}\right\rbrace:

= \ln\left(\frac{5}{4}\right) + \ln\left(\frac{6}{5}\right)^2 + \ln\left(\frac{4}{3}\right)^3 + \ln\left(k^8\right)

= \ln\left(\frac{5}{4} \cdot \left(\frac{6}{5}\right)^2 \cdot \left(\frac{4}{3}\right)^3 \cdot k^8\right)

Substituting k^2=\frac{e}{2}:

=\ln\left(\frac{5}{4}\cdot\frac{36}{25}\cdot\frac{64}{27}\cdot\frac{e^4}{16}\right)

The argument can be simplified:

\begin{align}\frac{5}{4}\cdot\frac{36}{25}\cdot\frac{64}{27}\cdot\frac{e^4}{16} & =\frac{36}{4}\cdot\frac{5}{25}\cdot\frac{64}{16}\cdot\frac{e^4}{27}\\ \placeholder{} & \placeholder{}\\ \placeholder{} & =9\cdot\frac15\cdot4\cdot\frac{e^4}{27}\\ \placeholder{} & \placeholder{}\\ \placeholder{} & =\frac{4e^4}{15}\end{align}

So,

\begin{align}\mathop{\mathrm{E}}(X)=\ln\left(\frac{4e^4}{15}\right) & =\ln\left(e^4\right)+\ln4-\ln15\\ \placeholder{} & \placeholder{}\\ \placeholder{} & =\boxed{4+\ln4-\ln15}\end{align}

x	1	2	3	4
P(X=x)	ln(45)	ln(56)	ln(34)	2lnk

x	1	2	3	4
P(X=x)	ln(45)	ln(56)	ln(34)	2lnk