I need to solve the exponential equation $$\frac{(x + 4)10^x}{x - 3} = 2x(10^x),$$assuming the fact that $2^x$ is always positive.
The example uses the case $x^3-2^x - 3(2^x) = 0$ factors out $2^x$ leaving $2^x (x^3-3) = 0$. They then apply the zero-product property and solve $2^x$ and $x^3 - 3 = 0$. Does this still apply for my more complex question? In that case, would I get one side of the equation to $0$, then apply this same property? Thanks in advance for any assistance.
\begin{align*} & \frac{(x+4)10^x-2x(x-3)10^x}{x-3}=0\\ \Rightarrow \quad & \frac{10^x(-2x^2+7x+4)}{x-3}=0\\ \Rightarrow \quad & -2x^2+7x+4=0 \quad \text{and} \quad x\neq 3 \end{align*}