Question 395078
First we use general exponential model equation and the initial information to find the specific exponential equation for this problem. In the general equation, the {{{A[0]}}} represents "the A when t = 0. The earliest value for A we have is from five years ago. So {{{A[0]}}} = 1400. This makes the other data, 1000 birds (A = 1000) five years later (t = 5) another point that should fit this equation. We now know 3 of the 4 unknowns in the general model. We can use the equation to find the 4th unknown, k:
{{{1000 = 1400*e^(k(5))}}}
First we divide by 1400:
{{{1000/1400 = e^(k(5))}}}
which simplifies to
{{{5/7 = e^(5k)}}}
Next we find the natural log of each side:
{{{ln(5/7) = ln(e^(5k))}}}
Then we use a property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, to move the exponent of the argument out in front. (It is this property that is the very reason we use logarithms on equations where the variable is in an exponent!)
{{{ln(5/7) = (5k)*ln(e)}}}
By definition, ln(e) = 1 so this becomes:
{{{ln(5/7) = 5k}}}
Dividing by 5 we get:
{{{ln(5/7)/5 = k}}}
This is an exact expression for k. This makes the specific equation to model this problem:
{{{A = 1400*e^((ln(5/7)/5)t)}}}<br>
For practical use we should replace the k with its decimal approximation (which we can find using our calculator):
{{{A = 1400*e^((-0.3364722366212129/5)t)}}}
{{{A = 1400*e^(-0.0672944473242426t)}}}<br>
We can now use this equation to find when the bird population drops below 100:
{{{99 = 1400*e^(-0.0672944473242426t)}}}
We solve this for t, just like we solved the earlier equation for k:
{{{99/1400 = e^(-0.0672944473242426t)}}}
{{{ln(99/1400) = ln(e^(-0.0672944473242426t))}}}
{{{ln(99/1400) = (-0.0672944473242426t)*ln(e)}}}
{{{ln(99/1400) = -0.0672944473242426t)}}}
{{{ln(99/1400)/(-0.0672944473242426) = t)}}}
{{{(-2.6491076654687601)/(-0.0672944473242426) = t)}}}
39.3659175578729793 = t
So the bird population will drop below 99 approximately 39.4 years after the t=0 time. Since the t=0 time was 5 years ago, the bird population will drop below 100 approximately 34.4 years from now.