population of bacteria after some years

Question

Suppose we have a population of bacterial that grows by 10 % every day. Suppose right now we have 10 000 bacteria count then in 10 days how many bacteria we got??

Attempt

Let $A_0 = 10000$ be initial amount. So, $A(1) = A_0 + A_0\frac{1}{10} $ and $A(2) = A_0 + \frac{1}{10} A_0 + \frac{1}{10} A_0 + \frac{1}{100} A_0 $. and so on... but after $10$ years the formula will get messy. Is it a simpler way to do this?

$A_n = (1.1)^n A_0$ so $\log A_n = n \log (1.1) + \log A_0$ ? — kimchi lover, Sep 14 '17 at 01:11
$(1+\frac 1{10})^{10}$ expands by the binomal theorem into $1 + \frac {10}{10} + \frac {45}{100} + \frac {120}{1000}+\frac {210}{10^4} + \frac {252}{10^6} + \cdots$ and beyond that you are looking at fractional bacteria, and that doesn't make a whole lot of sense.. — Doug M, Sep 14 '17 at 01:16
Neglecting the fractional bacteria seems to be a smaller mistake than assuming an unchecked 10% growth per day. — kimchi lover, Sep 14 '17 at 01:20

score 1 · Accepted Answer · answered Sep 14 '17 at 01:15

Call the population at the end of the $n$th day $P(n)$.

A $10 \%$ increase means multiplying the previous population by $1.1$

At the start, $P(0) = 10000$

At the end of day $1$, $P(1) = 1.1(10000)$

At the end of day $2$, $P(2) = (1.1)(1.1)(10000) = 1.1^2(10000)$

...

At the end of day $k$, $P(k) = 1.1^k(10000)$

Can you proceed?

population of bacteria after some years

Attempt

1 Answers1