# Updating half life

*20-Dec-2019 02:28*

If $b \gt 1$, then the population size doubles after a time of $T_=\frac$.If

If $x_= 2 x_$, then the doubling time is $T_=t_2-t_1$.

Also, at $T=70$, $P = 0.2$ so the population also doubled from 0.1 to 0.2 between $T=48$ and $T=70$, which is also 22 minutes.

\lt b \lt 1$, then the population size halves after a time of $T_ = \frac$.We'll show that this $T_$ won't depend on our choice of $t_1$.To determine $t_2$ (and hence $T_$), we must solve the equation \begin x_ = 2 x_ \end which, according to the model of equation \eqref, we can rewrite as \begin x_0 \times b^ = 2 x_0 \times b^.For the equation, $P_T = 0.022 \times 1.032^T$, the doubling time is $\log 2 / \log 1.032 = 22.0056$, as shown in the above applet.For the exponential equation $y_t = y_0 \times b^t$ with [[If $b \gt 1$, then the population size doubles after a time of $T_=\frac$.If $0 \lt b \lt 1$, then the population size halves after a time of $T_ = \frac$.We'll show that this $T_$ won't depend on our choice of $t_1$.To determine $t_2$ (and hence $T_$), we must solve the equation \begin x_ = 2 x_ \end which, according to the model of equation \eqref, we can rewrite as \begin x_0 \times b^ = 2 x_0 \times b^.For the equation, $P_T = 0.022 \times 1.032^T$, the doubling time is $\log 2 / \log 1.032 = 22.0056$, as shown in the above applet.For the exponential equation $y_t = y_0 \times b^t$ with $0 \lt b \lt 1$, the quantity $y_t$ does not grow with time $t$. The half-life, $T_$ is the time it takes for $y_t$ to decrease by one-half.\end Dividing both sides of the equation by $x_0 \times b^$, we can simplify the condition to \begin \frac = 2, \end which is the same as \begin b^ = 2.\end As we claimed at the beginning, we can find an equation for $T_=t_2-t_1$ that doesn't depend on our choice of $t_1$.

||If $b \gt 1$, then the population size doubles after a time of $T_=\frac$.

If $0 \lt b \lt 1$, then the population size halves after a time of $T_ = \frac$.

We'll show that this $T_$ won't depend on our choice of $t_1$.

To determine $t_2$ (and hence $T_$), we must solve the equation \begin x_ = 2 x_ \end which, according to the model of equation \eqref, we can rewrite as \begin x_0 \times b^ = 2 x_0 \times b^.

For the equation, $P_T = 0.022 \times 1.032^T$, the doubling time is $\log 2 / \log 1.032 = 22.0056$, as shown in the above applet.

For the exponential equation $y_t = y_0 \times b^t$ with $0 \lt b \lt 1$, the quantity $y_t$ does not grow with time $t$. The half-life, $T_$ is the time it takes for $y_t$ to decrease by one-half.

\end Dividing both sides of the equation by $x_0 \times b^$, we can simplify the condition to \begin \frac = 2, \end which is the same as \begin b^ = 2.

]] \lt b \lt 1$, the quantity $y_t$ does not grow with time $t$. The half-life, $T_$ is the time it takes for $y_t$ to decrease by one-half.\end Dividing both sides of the equation by $x_0 \times b^$, we can simplify the condition to \begin \frac = 2, \end which is the same as \begin b^ = 2.\end As we claimed at the beginning, we can find an equation for $T_=t_2-t_1$ that doesn't depend on our choice of $t_1$. [[If $x_= 2 x_$, then the doubling time is $T_=t_2-t_1$.

Also, at $T=70$, $P = 0.2$ so the population also doubled from 0.1 to 0.2 between $T=48$ and $T=70$, which is also 22 minutes.

||If $x_= 2 x_$, then the doubling time is $T_=t_2-t_1$.Also, at $T=70$, $P = 0.2$ so the population also doubled from 0.1 to 0.2 between $T=48$ and $T=70$, which is also 22 minutes.

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