Step $3$

Thus, $5=10e^{30k},$ or $e^{30k}=1/2$

Therefore, $k=\frac{ln(\frac{1}{2})}{30}$

$-\frac{ln(2)}{30}$

Stop $4$

Now, remembering that $ln(x^n)=nln(x)$ and that $e^{ln(z)}=z,$ we have the following.

$y(t)=10e^{(\frac{t}{30})(-ln(2))}=10*({)}^{-\frac{t}{30}}mg$