For exponential decay we have $\displaystyle g^{t_{1/2}}=\frac{1}{2}$. By taking the natural logarithm on both sides and using the calculation rules for logarithms you get:

$$\ln(g^{t_{1/2}})=\ln(\frac{1}{2})$$ Thus $$t_{1/2} \cdot \ln(g) = -\ln(2)$$ After rewriting: $$t_{1/2} = -\frac{\ln(2)}{\ln(g)}$$ Because $g=e^r$, we can also write $\ln(g)=r$ and $$t_{1/2} = -\frac{\ln(2)}{r}\approx -\frac{0.693}{r}$$

For exponential growth holds $g^{t_{d}}=2$. By taking the natural logarithm on both sides and using the calculation rules for logarithms you get:

$$t_{d} = \frac{\ln(2)}{\ln(g)}$$ Because $g=e^r$, we can also write $$t_{d} = \frac{\ln(2)}{r}\approx \frac{0.693}{r}$$