Question

1. during an experiment it is recorded that there are 35 bacteria present originally in a culture. after 3 h there are 17920 bacteria. what is the doubling period? leave your answer in fraction. a = a₀(2)^(t/d) g: = 17920 a₀ = 35 t = 3 r: d 512=(2)^(3/d)

Explanation:

Step1: Substitute values into formula

The formula for exponential - growth with doubling is $A = A_0(2)^{\frac{t}{d}}$, where $A$ is the final amount, $A_0$ is the initial amount, $t$ is the time elapsed, and $d$ is the doubling - period. Given $A = 17920$, $A_0=35$, and $t = 3$. Substitute these values: $17920=35\times(2)^{\frac{3}{d}}$. First, divide both sides by 35: $\frac{17920}{35}=(2)^{\frac{3}{d}}$. Since $\frac{17920}{35}=512$, the equation becomes $512=(2)^{\frac{3}{d}}$.

Step2: Rewrite 512 as a power of 2

We know that $512 = 2^9$. So, $2^9=(2)^{\frac{3}{d}}$.

Step3: Equate the exponents

Since the bases are the same (both are 2), we can set the exponents equal to each other: $9=\frac{3}{d}$.

Step4: Solve for d

Cross - multiply to get $9d = 3$. Then, divide both sides by 9: $d=\frac{3}{9}=\frac{1}{3}$.

Answer:

$\frac{1}{3}$