To solve the equation you've provided, let's first rewrite it clearly:

\[
3^x + 2a = 3^x + 2 - 5
\]
\[
3^x = 36
\]

Starting with the second equation:

\[
3^x = 36
\]

We can solve for \( x \) by expressing \( 36 \) as a power of \( 3 \):

\[
36 = 3^2 \cdot 4
\]

Taking the logarithm base \( 3 \):

\[
x = \log_3(36) = \log_3(3^2) + \log_3(4) = 2 + \log_3(4)
\]

Now since we need to simplify this further and since this doesn’t yield an integer value directly, let's also isolate \( a \) by using the first equation.

Substituting \( x \) into the adjusted first equation:

\[
3^x + 2a = 3^x + 2 - 5
\]

This simplifies to:

\[
2a = 2 - 5
\]
\[
2a = -3
\]
\[
a = -\frac{3}{2}
\]

We now have \( a = -\frac{3}{2} \) and \( x \) as \( \log_3(36) \).

If you have any specific values you need or further questions, feel free to ask!