Let's simplify the equation step by step. The original equation given is:

\[
\frac{1}{2} \cdot 2^N + 4 \cdot 2^N = 9 \cdot 5^N
\]

First, we can factor out \(2^N\) from the left side:

\[
2^N \left( \frac{1}{2} + 4 \right) = 9 \cdot 5^N
\]

Now, we simplify the term in the parentheses:

\[
\frac{1}{2} + 4 = \frac{1}{2} + \frac{8}{2} = \frac{9}{2}
\]

So the equation becomes:

\[
2^N \cdot \frac{9}{2} = 9 \cdot 5^N
\]

Now, we can divide both sides by \(9\) (assuming \(N\) is such that \(9 \neq 0\)):

\[
\frac{2^N}{2} = 5^N
\]

This simplifies to:

\[
2^{N-1} = 5^N
\]

To solve for \(N\), we will take the logarithm of both sides. Using base 10 logarithms for convenience:

\[
\log(2^{N-1}) = \log(5^N)
\]

This can be rewritten using the properties of logarithms:

\[
(N-1) \log(2) = N \log(5)
\]

Now, distribute and rearrange the terms to isolate \(N\):

\[
N \log(2) - \log(2) = N \log(5)
\]

Bringing all \(N\) terms to one side:

\[
N \log(2) - N \log(5) = \log(2)
\]

Factor out \(N\):

\[
N (\log(2) - \log(5)) = \log(2)
\]

Now solve for \(N\):

\[
N = \frac{\log(2)}{\log(2) - \log(5)}
\]

Using the property of logarithms, we can express this as:

\[
N = \frac{\log(2)}{\log\left(\frac{2}{5}\right)}
\]

This gives us the exact solution for \(N\). If you need a decimal approximation, you may use a calculator to evaluate this expression.