How can I solve $5^{2x} + 4(5^x) - 5 = 0$? This is a math problem I'm currently working on.
$$5^{2x} + 4(5^x) - 5 = 0$$
I've used logarithm to try solve the problem. Here's what I've done so far: 
\begin{align}5^{2x} + 4(5^x) - 5 &= 0\\
5^{2x} + 5^x &= \frac{5}{4}\\
\log_5{2x} + \log_5{x} &= \log_5\left(\frac{5}{4}\right)\\
\log_5{2x^2} &= \log_5\left(\frac{5}{4}\right)\\
5^{2x^2} &= 5^\frac{5}{4}\\
(2x^2)\log5 &= \frac{5}{4}\log5\\
2x^2 &= \frac{5}{4}\\
2x^2 &= 1.25\\
x^2 &= 0.625\\
\sqrt {x^2} &= \sqrt {0.625}\\
x &= \frac{\sqrt {10}}{4}\end{align}
Substituting the value for $x$ into the equation doesn't equate it to zero. I've tried several different ways but I have still not come up with a correct answer. 

You made a pretty big mistake because $\log a + \log b \neq \log (a+b)$. So, your third line does not follow from your second.
To actually solve the problem, here's two hints:


*

*Set $y=5^x$

*Use the fact that $a^{bc} = (a^b)^c$

