To solve the equation

\[
6^x + 6^{x+1} = 2^{x+1} + 2 \cdot 2^{x+2} + 4 \cdot 2^x
\]

let's first simplify the left-hand side:

\[
6^x + 6^{x+1} = 6^x + 6 \cdot 6^x = 6^x(1 + 6) = 7 \cdot 6^x
\]

Now, simplifying the right-hand side:

\[
2^{x+1} + 2 \cdot 2^{x+2} + 4 \cdot 2^x
\]
- The first term is \(2^{x+1}\).
- The second term becomes \(2 \cdot 2^{x+2} = 2^{x+2+1} = 2^{x+3}\).
- The third term is \(4 \cdot 2^x = 2^2 \cdot 2^x = 2^{x+2}\).

So the right-hand side becomes

\[
2^{x+1} + 2^{x+3} + 2^{x+2}.
\]

Now combine these:

- \(2^{x+3} + 2^{x+2} + 2^{x+1} = 2^{x+1}(1 + 2 + 4) = 7 \cdot 2^{x+1}.\)

So we have the simplified equation:

\[
7 \cdot 6^x = 7 \cdot 2^{x+1}.
\]

We can divide both sides by \(7\) (since \(7 \neq 0\)):

\[
6^x = 2^{x+1}.
\]

Rearranging this gives:

\[
6^x = 2 \cdot 2^x.
\]

We can write \(6\) as \(2 \cdot 3\):

\[
(2 \cdot 3)^x = 2^{x + 1}.
\]

Expanding the left-hand side, we get:

\[
2^x \cdot 3^x = 2^{x + 1}.
\]

Dividing both sides by \(2^x\) gives:

\[
3^x = 2.
\]

Taking the logarithm of both sides:

\[
x \log(3) = \log(2),
\]

so

\[
x = \frac{\log(2)}{\log(3)}.
\]

Thus, the solution to the equation is

\[
\boxed{\frac{\log(2)}{\log(3)}}.
\]