Rewrite $\frac{125}{\left(\frac{1}{625}\right)^{-x-3}}=5^3$ in a common base then solve for $x$ I am to rewrite $\frac{125}{(\frac{1}{625})^{-x-3}}=5^3$ and then solve for x. My textbooks solutions section says the solution is -3. I gave it a shot and got 3.25. Here is my working:
$$\frac{125}{\left(\frac{1}{625}\right)^{-x-3}}=5^3$$
$$\frac{5^3}{\left(\frac{1}{5^4}\right)^{-x-3}}=5^3$$
(Here's where I get pretty confused. I think I can pull the denominator component $\frac{1}{5^4}$ up above and transform from a fraction to a full number, but I'm unsure how to handle the exponent $-x-3$)
$$5^3 \times 5^{4(x-3)}=5^3$$
$$5^{12x-36}=5^3$$
$$12x-36=3$$
$$12x=39$$
$$x=\frac{39}{12}=3.25$$
Where did I go wrong and how can I arrive at 3? I suspect I went stray between the second and 3rd lines. If this is so, rather than just provide a solution please please do also explain the logic explicitly... how to I 'pull up' $\left(\frac{1}{5^4}\right)^{-x-3}$, what are the rules for each component?

Your first mistake starts here:

$$5^3 \times 5^{4(x-3)}=5^3$$

This is not correct. Because,
$$\begin{align}\left(\frac{1}{625}\right)^{-x-3}&=\left(5^{-4}\right)^{-x-3}\\
&=5^{-4\times (-x-3)}\\
&=5^{4(x+3)}.\end{align}$$
Your second mistake is:

$$5^3 \times 5^{4(x-3)}=5^3$$
$$5^{12x-36}=5^3$$

The correct one is as follows:
$$5^{3+4(x-3)}=5^3$$
But, we don't need this.  Because there is a simpler way:
$$\begin{align}5^3 \times 5^{4(x-3)}=5^3 &\iff 5^{4(x-3)}=5^0\\ &\iff 4(x-3)=0\end{align}$$

However, even the last one we wrote is not valid. Because, we have
$$\begin{align}5^3 \times 5^{4(x+3)}=5^3 &\iff 5^{4(x+3)}=5^0\\ &\iff 4(x+3)=0 \\ &\iff x=-3.\end{align}$$
