modeling real life the length ℓ (in centimeters) of a scalloped hammerhead shark can be modeled by the…

modeling real life the length ℓ (in centimeters) of a scalloped hammerhead shark can be modeled by the function ℓ = 266 - 219e^(-0.05t) where t is the age (in years) of the shark. how old is a shark that is 175 centimeters long? round your answer to the nearest tenth. the shark is about □ years old.

modeling real life the length ℓ (in centimeters) of a scalloped hammerhead shark can be modeled by the function ℓ = 266 - 219e^(-0.05t) where t is the age (in years) of the shark. how old is a shark that is 175 centimeters long? round your answer to the nearest tenth. the shark is about □ years old.

Answer

Explanation:

Step1: Substitute length value

Given $\ell = 175$ and $\ell=266 - 219e^{- 0.05t}$, we substitute $\ell$: $175=266 - 219e^{-0.05t}$

Step2: Rearrange the equation

First, move the terms around: $219e^{-0.05t}=266 - 175$ $219e^{-0.05t}=91$ Then, $e^{-0.05t}=\frac{91}{219}$

Step3: Take natural - logarithm of both sides

$\ln(e^{-0.05t})=\ln(\frac{91}{219})$ Using the property $\ln(e^{x}) = x$, we get $-0.05t=\ln(\frac{91}{219})$

Step4: Solve for $t$

$t=\frac{\ln(\frac{91}{219})}{- 0.05}$ $t=\frac{\ln(91)-\ln(219)}{-0.05}$ $\ln(91)\approx4.5108$ and $\ln(219)\approx5.4905$ $t=\frac{4.5108 - 5.4905}{-0.05}=\frac{-0.9797}{-0.05}=19.6$

Answer:

$19.6$