Question

The number of a particular fish in thousands x years after 1972 can be modeled by f(x)=230(0.886)x. Estimate the year when the number of fish reached 99 thousand.

Answer 1

Answer:

The year is 1979.

Step-by-step explanation:

Given:

The function relating fish population and number of years since 1972 is:

$f(x)=230(0.886)^x\\Where,\ f(x)\rightarrow \textrm{fish population}\\x\rightarrow \textrm{number of years passed after 1972}$

Now, when the fish population reaches 99 thousand, it means that $f(x)=99$. Now, plugging in $f(x)=99$ in the above equation and solving for $x$. This gives,

$99=230(0.886)^x\\\frac{99}{230}=0.886^x\\\\\textrm{Taking natural log on both sides}\\\ln(0.886)^x=\ln(\frac{99}{230})\\\\\textrm{Using log property: }\log a^b=b\log a\\x\ln (0.886)=\ln(0.43)\\x\times (-0.12)=-0.84\\x=\frac{-0.84}{-0.12}=7\ years$

Therefore, the year after 7 years passed since 1972 is given as:

1972 + 7 = 1979.

So, the year when the number of fish reaches 99 thousand is 1979.