^{0.035t}

^{0}= that number (i.e. e

^{0}= 1)

^{(0.035)(11 years)}]

**n(t) = 98.5 million.**

What was the rat population in 1994?

What is the rat population going to be in the year 2005?

Tutors, sign in to answer this question.

Marked as Best Answer

n(t) = 67e^{0.035t}

In 1994 the rat population was 67,000,000 because any number times e^{0} = that number (i.e. e^{0} = 1)

The year 2005 was 11 years after 1994, so t = 11 years.

n(t) = 67*[e^{(0.035)(11 years)}]

Assuming the initial rat population was 67 million, simply plug in 11 for t, so:

2005-1994 = 11; Eleven years have passed.

0.035 * t = 0.035 * 11 = 0.385;

n(11) = 67 * e^{(0.035*11)}

n(11) = 67 * e^{0.385 }= 67 * 1.46961 = 98.4641 (million)

1.) In windows calculator, use View>Scientific view

2.) For e^{x }use [Inv]. This changes the [ln] button to [e^{x}].

3.) Multiply by 67.

.385 [Inv][e^{x}] or [Inv][ln] or [Inv][log] on your calculator, then times 67.

e = 2.7182... > 1. So with a positive exponent, the result will also be >1. The more years pass, the greater the exponent,and the greater the result (and number of rats) will be. When you multiply by a number greater than one, the result will always be a bigger number.

In year 1994, n(1) = 67e

In year 2005, n(12) = 67e

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.