How on earth do I solve? Any help will be much appreciated.
The value of $M$ is given by $M = a \log_{10}S + b$.
Note: Seismic moment measure the energy of the earthquake.
Using the following information, determine the values of $a$ and $b$ and hence find the seismic moment $(S)$ of an earthquake which has a magnetic moment $(M)$ of $7.9$.
Seismic Moment $(S) = 4.47 \times 10^{25}$, Magnetic Moment $(M) = 7$
Seismic Moment $(S) = 2 \times 10^{27}$, Magnetic Moment $(M) = 7.5$
You have \begin{align} M=a\log_{10}S+b \end{align} Then, from the given values you have \begin{align} 7=a\log_{10}(4.47\times10^{25})+b\tag1 \end{align} and \begin{align} 7.5=a\log_{10}(2\times10^{27})+b\tag2 \end{align} Subtract $(1)$ from $(2)$ yield \begin{align} 7.5-7&=a\log_{10}(2\times10^{27})+b-(a\log_{10}(4.47\times10^{25})+b)\\ 0.5&=a\log_{10}(2\times10^{27})-a\log_{10}(4.47\times10^{25})\\ 0.5&=a(\log_{10}(2\times10^{27})-\log_{10}(4.47\times10^{25}))\\ 0.5&=a\log_{10}\left(\frac{2\times10^{27}}{4.47\times10^{25}}\right)\\ a&=\frac{0.5}{\log_{10}\left(\frac{2\times10^{27}}{4.47\times10^{25}}\right)} \end{align} The value of $b$ can be obtained by substituting $a$ to $(1)$ or $(2)$.
\begin{align} 7&=\frac{0.5}{\log_{10}\left(\frac{2\times10^{27}}{4.47\times10^{25}}\right)}\log_{10}(4.47\times10^{25})+b\\ b&=7-\frac{0.5\log_{10}(4.47\times10^{25})}{\log_{10}\left(\frac{2\times10^{27}}{4.47\times10^{25}}\right)} \end{align}