Answer
$E_{1}=10^{3R_{1}/2+11.8}$
$E_{2}=10^{3R_{2}/2+11.8}$
$\displaystyle \frac{E_{2}}{E_{1}}=\frac{10^{3R_{2}/2+11.8}}{10^{3R_{1}/2+11.8}}\qquad $(use the rule $\displaystyle \frac{a^{m}}{a^{n}}=a^{m-n} )$
$=10^{3R_{2}/2+11.8 - (3R_{1}/2+11.8)}$
$=10^{1.5R_{2}+11.8 - 1.5R_{1}-11.8)}$
$=10^{1.5(R_{2}-R_{1})}$
Work Step by Step
No extra steps. The exercise asked to show that the relation
$\displaystyle \frac{E_{2}}{E_{1}}=10^{1.5(R_{2}-R_{1})}$
is valid.
