Answer: C. About 6 days
Step-by-step explanation:
Here the function that shows the population of fish after x weeks,
[tex]P(x) = 6^{bx}[/tex]
Where b is any unknown,
If b = 3,
Then, the function is,
[tex]P(x) = 6^{3x}[/tex]
Which is a exponentially increasing function,
That having y-intercept = (0,1)
And, horizontal asymptote,
y = 0
End behavior of the function:
As [tex]x\rightarrow \infty[/tex] , [tex]y\rightarrow \infty[/tex]
As [tex]x\rightarrow -\infty[/tex] , [tex]y\rightarrow 0[/tex]
Thus, by the above information we can graph the given relation.
Now, For at least 100 fish in the pound,
[tex]6^{3x}\geq 100[/tex]
By taking log on both sides,
[tex] 3x log(6) \geq log(100)[/tex]
[tex]3x\geq \frac{log(100)}{log(6)}[/tex]
[tex]3x\geq 2.57019441788[/tex]
[tex]x\geq 0.85673147262\approx 0.8567[/tex]
Thus, after 0.8567 weeks (approx) the fish on the pound will be at least 100.
1 week = 7 days,
0.8567 weeks = 5.9969 days ≈ 6 days
Hence, after 6 days (approx) the fish on the pound will be at least 100.
