Solve the equation: $ 3^{2x+1}+4 \cdot 3^x = 15 $ Solve the equation: $ 3^{2x+1}+4 \cdot 3^x = 15 $ where $x$ is a real number.
Background: Doing Olympiad question and got one from the book.
Attempt:
Let $3^x$ be $u$.
\begin{align*}
3^{2x+1} + 4 \cdot 3x - 15 &= 0 \\
3^{2x} \cdot 3^1 + 4 \cdot 3^x - 15 &= 0 \\
3^{2x} \cdot 3 + 4 \cdot 3^x - 15 &= 0 \\
3 \cdot 3^{2x} + 4 \cdot 3^x - 15 &= 0 \\
3u^2 + 4u - 15 &= 0
\end{align*}
Factoring the equation we get
$$ u=\frac{5}{3} $$
(We eliminate $u=-3$ as $x$ is real.)
$$ 3^x=\frac{5}{3} $$
Taking $\log$ both sides
$$ x \log 3 = \log\left(\frac{5}{3}\right) $$
Now I want to know how to further solve this. Also is there any easier way to solve this?

It depends. It seems as if you already have a short solution and you may consider taking base-$3$ logarithm to make a shorter one, but most of the time, logarithms are of base-$e$, which is exactly what you did.
