Multiplying Fuzzies Problem

First, we'll need to make a table like this:

Note : Let $f$ represent the number of fuzzies at a given time.

We're given that the starting number of Fuzzies is 1, so we know that when $t = 0$ , $f = 1$ .

We also know that, since $f$ doubles as $t$ increases, the graph will be exponential. In other words, the formula will be of this form:

$y=e^{ax}$

Or, with our variables:

$f = e^{at}$

With $a$ being a constant.

So, with this equation, we need to solve for the time at which $f$ reaches the max capacity (75008).

$75008=e^{at}$

$ln(75008)=at$

$11.225=at$

Now, all we need to do is find $a$ . We already know that $f=2$ when $t=1$ , so solving the exponential equation for $a$ (using natural log as shown) gives that $a$ is equal to $ln(2)=0.6931$

So, now that we know $a$ , we can easily solve for $t$ .

$\frac{11.225}{0.6931}=t$

$t=16.194 \text{ s}$

So, the nearest whole second before $f$ reaches 75008 is the 16th second.

$2^x<75008$ . $16$ is the biggest integer that can make this inequality true, so the answer is $\boxed{\color{#D61F06}the \ 16th \ second}$ .