$$\begin{array}{rll}
30000&=&15500(1+\frac{5.75}{4})^{4t}\\\\
\frac{30000}{15500}&=&(1+\frac{5.75}{4})^{4t}\\\\
\frac{60}{31}&=&(2.4375)^{4t}\\\\
log\left(\frac{60}{31}\right)&=&log\left(2.4375^{4t}\right)\\\\
log\left(\frac{60}{31}\right)&=&4t\;log \;2.4375\\\\
\dfrac{log\left(\frac{60}{31}\right)}{4log \;2.4375}&=&t\\\\
\end{array}$$
$${\frac{{log}_{10}\left({\frac{{\mathtt{60}}}{{\mathtt{31}}}}\right)}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{2.437\: \!5}}\right)\right)}} = {\mathtt{0.185\: \!291\: \!084\: \!619\: \!587\: \!8}}$$
.$$\begin{array}{rll}
30000&=&15500(1+\frac{5.75}{4})^{4t}\\\\
\frac{30000}{15500}&=&(1+\frac{5.75}{4})^{4t}\\\\
\frac{60}{31}&=&(2.4375)^{4t}\\\\
log\left(\frac{60}{31}\right)&=&log\left(2.4375^{4t}\right)\\\\
log\left(\frac{60}{31}\right)&=&4t\;log \;2.4375\\\\
\dfrac{log\left(\frac{60}{31}\right)}{4log \;2.4375}&=&t\\\\
\end{array}$$
$${\frac{{log}_{10}\left({\frac{{\mathtt{60}}}{{\mathtt{31}}}}\right)}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{2.437\: \!5}}\right)\right)}} = {\mathtt{0.185\: \!291\: \!084\: \!619\: \!587\: \!8}}$$