Doubling time is the amount of time it takes for a given quantity to double in size or value at a constant growth rate. This means that the ratio of the final and initial values of the exponential function will be equal to 2:
[tex]\begin{gathered} \text{For} \\ y=a(b)^x^{} \\ We\text{ have} \\ \frac{y}{a}=2 \end{gathered}[/tex]
The function is given to be:
[tex]y=At^{\rho}[/tex]
For the doubling time, we have that:
[tex]\frac{y}{A}=2[/tex]
If we divide both sides of the equation by A, we have:
[tex]\frac{y}{A}=t^{\rho}[/tex]
Substituting for the ratio, we have:
[tex]2=t^{\rho}[/tex]
Finding the natural logarithm of both sides, we have:
[tex]\ln 2=\ln t^{\rho}[/tex]
Applying the law of exponents given to be:
[tex]\ln a^b=b\ln a[/tex]
we have that:
[tex]\ln 2=\rho\ln t[/tex]
Divide both sides by ln(t), we have:
[tex]\rho=\frac{\ln2}{\ln t}[/tex]
Since we have:
[tex]\frac{y}{A}=2[/tex]
We can have the time to be:
[tex]\begin{gathered} \rho=\frac{\ln (\frac{y}{A})}{\ln t} \\ \text{Given} \\ \frac{y}{A}=2 \end{gathered}[/tex]
