<s>[INST] 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?
 [/INST] HINT:
You want to find value of $x$
It can be proceeded as
$x\log3=\log(\frac{5}{3})$
$x=\frac{\log(\frac{5}{3})}{\log3}$
$x=\log_3(\frac{5}{3})$ by converse of base change theorem
$x=\log_3 5-1$
To make your solution short by two or three steps take $\log_3()$ on both sides
$3^x=5/3$
$\log_3 3^x=\log_3 (5/3)$
$x=\log_3(5/3)$
$x=\frac{\log 5 - \log 3}{\log3}$
$x=\frac{0.698 - 0.477}{0.477}$
$x=\frac{0.221}{0.477}$ which can nearly be approximated to $0.5$
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