College Algebra, Chapter 5, 5.3, Section 5.3, Problem 68

Estimate a student's score on a Biology test two years after he got a score of 80 on a test covering the same material, by using Ebbinghaus' Law of Forgetting. Assume that $c= 0.3$ and $t$ is measured in months.

$\log P = \log = P_0 - c \log (t + 1) \qquad$ Model

Solving for $P$


$
\begin{equation}
\begin{aligned}

\log P =& \log P_0 - \log (t + 1)^2
&& \text{Law of Logarithm } \log_a (A^C) = C \log_a A
\\
\\
\log P =& \log \frac{P_0}{(t + 1)^2}
&& \text{Law of Logarithm } \log_a \left( \frac{A}{B} \right) = \log_a A - \log_a B
\\
\\
P =& \frac{P_0}{(t + 1)^2}
&& \text{Because $\log$ is one-to-one}

\end{aligned}
\end{equation}
$



Here $P_0 = 80, C = 0.3$ and $t$ is measured in months

In two years: $t = 24$ months

and


$
\begin{equation}
\begin{aligned}

P =& \frac{80}{(24 + 1)^{0.3}}
\\
\\
P =& 30.46

\end{aligned}
\end{equation}
$



The student's expected score on Biology test after 2 years is $30.46$.