Polynomial Evaluation 2
Let $ P(x) $ be the degree three polynomial with
\begin{align*}
P(1) &= \log 1, \\
P(2) &= \log 2, \\
P(3) &= \log 3, \\
P(4) &= \log 4.\end{align*}Then $ P(5) $ can be expressed in the form $ A \log \frac{B}{C}, $ where $ A, $ $ B, $ and $ C $ are positive integers, and $ C $ is prime. Find $ A + B + C $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Let the cubic polynomial be $ P(x) = ax^3 + bx^2 + cx + d $. Then
\begin{align*}
a + b + c + d &= P(1), \\
8a + 4b + 2c + d &= P(2), \\
27a + 9b + 3c + d &= P(3), \\
64a + 16b + 4c + d &= P(4), \\
125a + 25b + 5c + d &= P(5).\end{align*}Subtracting the first and second equations, second and third equations, and third and fourth equations, we get
\begin{align*}
7a + 3b + c &= P(2) - P(1), \\
19a + 5b + c &= P(3) - P(2), \\
37a + 7b + c &= P(4) - P(3), \\
61a + 9b + c &= P(5) - P(4).\end{align*}Again subtracting the equations in pairs, we get
\begin{align*}
12a + 2b &= P(3) - 2P(2) + P(1), \\
18a + 2b &= P(4) - 2P(3) + P(2), \\
24a + 2b &= P(5) - 2P(4) + P(3).\end{align*}Then
\begin{align*}
6a &= P(4) - 3P(3) + 3P(2) - P(1), \\
6a &= P(5) - 3P(4) + 3P(3) - P(2),
\end{align*}so $ P(5) - 3P(4) + 3P(3) - P(2) = P(4) - 3P(3) + 3P(2) - P(1) $.
Hence,
\begin{align*}
P(5) &= 4P(4) - 6P(3) + 4P(2) - P(1) \\
&= 4 \log 4 - 6 \log 3 + 4 \log 2 - \log 1 \\
&= 4 \log 2^2 - 6 \log 3 + 4 \log 2 \\
&= 8 \log 2 - 6 \log 3 + 4 \log 2 \\
&= 12 \log 2 - 6 \log 3 \\
&= 6 \log 4 - 6 \log 3 \\
&= 6 \log \frac{4}{3}.\end{align*}Therefore, $ A + B + C = 6 + 4 + 3 = \boxed{13} $.
Freeflows