The population of a culture of bacteria is modeled by the logistic equation:
$P(t)\;=\;\frac{14,250}{1+29\cdot e^{-0.62t}}$
To the nearest tenth, how many days will it take the culture to reach 75% of its carrying capacity?
What is the carrying capacity?
What is the initial population for the model?
$P(0)\;=\;\frac{14,250}{1+29\cdot e^{-0.62\cdot0}}\\P(0)\;=\;\frac{14,250}{30}=475\\P(0)\;=\;P_0\;=\;475\\$
Now, we got $P_0$.
75% is $\frac{3}{4}$.
So,
$P(t)\;=\;P_0e^{k\cdot t}\\\frac34\;=\;475\cdot e^{k\cdot t}$
I'm stuck at finding $k$.
Not sure I'm approaching in the right way.
The carrying capacity is the maximum value of $P(x)$. Maximising it requires minimising the denominator, which is $$1+29e^{-0.62t} $$ and it is achieved when $t\to \infty$, and the denominator goes to $1$. The carrying capacity hence is $$14250$$ Now, you need to find $t$ for which $P(t)=75\%$ of $14250$, i.e. $$\frac 34 \times 14250 =\frac{14250}{1+29e^{-0.62 t}} \\ \implies29e^{-0.62t}=\frac 13 \\ \implies e^{0.62t}=87 \\ \implies t=\frac{\ln 87}{0.62}\approx 7.2$$