Consider the following function $g(x)$ defined as\[(x^{2^{2008}-1}-1)g(x) = (x+1)(x^2+1)(x^4+1)\cdots (x^{2^{2007}}+1) - 1\]Find $g(2)$.

Multiply both sides by $x-1$; the right hand side collapses by the reverse of the difference of squares.
\begin{align*}(x-1)(x^{2^{2008}-1}-1)g(x) &= (x-1)(x+1)(x^2+1)(x^4+1)\cdots (x^{2^{2007}}+1) - (x-1)\\ &= (x^2-1) (x^2+1)(x^4+1)\cdots (x^{2^{2007}}+1) - (x-1)\\ &= \cdots\\ &= \left(x^{2^{2008}}-1\right) - (x-1) = x^{2^{2008}} - x \end{align*}Substituting $x = 2$, we have\[\left(2^{2^{2008}-1}-1\right) \cdot g(2) = 2^{2^{2008}}-2 = 2\left(2^{2^{2008}-1}-1\right)\]Dividing both sides by $2^{2^{2008}-1}$, we find $g(2) = \boxed{2}$.
2
Level 5
Intermediate Algebra
Problem: Consider the following function $g(x)$ defined as\[(x^{2^{2008}-1}-1)g(x) = (x+1)(x^2+1)(x^4+1)\cdots (x^{2^{2007}}+1) - 1\]Find $g(2)$.

Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find \( g(2) \), we first need to evaluate the left-hand side of the equation at \( x = 2 \). We substitute \( x = 2 \) into the equation:

\[
(2^{2^{2008}-1}-1)g(2) = (2+1)(2^2+1)(2^4+1)\cdots (2^{2^{2007}}+1) - 1
\]

Calculating the right-hand side, we start with the product:

\[
(2+1) = 3, \quad (2^2+1) = 5, \quad (2^4+1) = 17, \quad (2^8+1) = 257, \ldots, (2^{2^{2007}}+1)
\]

This product can be expressed as:

\[
\prod_{k=0}^{2007} (2^{2^k} + 1)
\]

Next, we recognize that the product \( (x^{2^n} - 1) \) can be factored into \( (x-1)(x+1)(x^2+1)\cdots(x^{2^{n-1}}+1) \). For \( x = 2 \), we have:

\[
2^{2^{2008}} - 1 = (2-1)(2+1)(2^2+1)(2^4+1)\cdots(2^{2^{2007}}+1)
\]

Thus, we can relate the product to \( 2^{2^{2008}} - 1 \):

\[
(2^{2^{2008}} - 1) = 1 \cdot 3 \cdot 5 \cdots \cdots \cdots \cdots (2^{2^{2007}} + 1)
\]

Now, we can compute the left-hand side:

\[
(2^{2^{2008}-1} - 1)g(2)
\]

This means we need to evaluate \( 2^{2^{2008}-1} - 1 \). The left-hand side simplifies to:

\[
(2^{2^{2008}-1} - 1)g(2) = \prod_{k=0}^{2007} (2^{2^k} + 1) - 1
\]

To isolate \( g(2) \), we rearrange:

\[
g(2) = \frac{\prod_{k=0}^{2007} (2^{2^k} + 1) - 1}{2^{2^{2008}-1} - 1}
\]

Next, we evaluate the product \( \prod_{k=0}^{2007} (2^{2^k} + 1) \). This product is known to equal \( 2^{2^{2008}} - 1 \). Therefore, we have:

\[
g(2) = \frac{(2^{2^{2008}} - 1) - 1}{2^{2^{2008}-1} - 1}
\]

This simplifies to:

\[
g(2) = \frac{2^{2^{2008}} - 2}{2^{2^{2008}-1} - 1}
\]

Factoring out \( 2 \) from the numerator gives:

\[
g(2) = \frac{2(2^{2^{2008}-1} - 1)}{2^{2^{2008}-1} - 1}
\]

This simplifies to:

\[
g(2) = 2
\]

Thus, the final answer is \( 2 \).<|END_OF_LATENT|> The final answer is $2$.