An algebra problem by Aira Thalca

Starting with $25^x = 10^x + 4^{x+1}$ , we divide both sides by $4^x$

Now, we have $\begin{aligned}\begin{aligned}\left(\frac{25}{4}\right)^x&=&\left(\frac{5}{2}\right)^x+4\\\left(\frac{5}{2}\right)^{2x}-\left(\frac{5}{2}\right)^x-4&=&0\end{aligned}\end{aligned}$

Let $a=\left(\dfrac{5}{2}\right)^x$ . This makes the equation become $a^2-a-4= 0 \longrightarrow a = \dfrac{1\pm\sqrt{17}}{2}$ .

Since $(\frac{5}{2})^{2x}$ and $(\frac{5}{2})^x$ are positive numbers, the value of $(\frac{5}{2})^x$ is $\dfrac{1 + \sqrt{17}}{2}$ .

So, the value of $x$ is $x=\boxed{\log_{\frac52}{\left(\dfrac{1+\sqrt17}2\right)}}$

So, the value of $abcde$ is $5 • 2 • 1 • 17 • 2 = 340$