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?
 A: 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
$$\begin{align}5^3 \times 5^{4(x+3)}=5^3 &\iff 5^{4(x+3)}=5^0\\ &\iff 4(x+3)=0 \\ &\iff x=-3.\end{align}$$
A: $\frac{5^3}{\left(\frac{1}{5^4}\right)^{\color{red}{\large -}x-3}}=5^3$
Here you have $\frac{5^3}{\left(\frac{1}{5^4}\right)^{\color{red}{\large\text{negative sign}}x-3}}=5^3$
Here you correctly surmised that we can "pull" the $\frac 1{5^4}$ "up" to get that $\frac 1{(\frac 1{5^4})^M} = (5^4)^M$.  That is good.  So you wrote
$5^3 \times 5^{4(\color{green}{\large \text{NO NEGATIVE SIGN}}x-3)}=5^3$
What happened to the negative sign before the $x$?
You ought to have $5^3 \times 5^{4(-x-3)} = 5^3$
Your second mistake is you figured:
$5^3 \times 5^{4(\color{green}{\large \text{NO NEGATIVE SIGN}}x-3)}= 5^{3\color{red}{\large\times} 4(x-3)}$
It should be $5^3 \times 5^{4(-x-3)} = 5^{3 \color{green}{\large +} 4(-x-3)}$.
From there you'd get everything right.
$5^{3 \color{green}{\large +} 4(-x-3)}= 5^3$
$5^{-4x - 12 + 3} = 5^3$
$5^{-4x -9} = 5^3$
$-4x -9 = 3$
$-4x = 12$
$x = \frac {12}{-4} = -3$.
But note:  We didn't have to expand $3+4(-x-3)$ out to $-4x -9$.  We could have just left it as is and get $3+4(-x-3) = 3$ and subtract three from both sides to get $4(-x-3) = 0$ to get $x+3 = 0$ and $x =-3$.
Also from $\frac{5^3}{\left(\frac{1}{5^4}\right)^{\color{red}{\large -}x-3}}=5^3$ we could have divided both sides byt $5^3$ to get $\frac 1{(\frac 1{5^4})^{-x-3}} = 1$
Also if we are being more systematic we could have done either:
$\frac {5^3}{(\frac 1{5^4})^{-x-3}} = \frac {5^3}{(5^{-4})^{-x-3}}=\frac {5^3}{5^{-4(-x-3)}}=\frac {5^3}{5^{4(x+3)}} = 5^{3- 4(x+3)}$
or
$\frac {5^3}{(\frac 1{5^4})^{-x-3}} = 5^3 \times (\frac 1{5^4})^{x+3} =5^3 \times 5^{-4(x+3)}$
There's a plethora of options.
A: Since $5^3=125$ we have
$$\frac{1}{\text{(some non-zero number)}^{-x-3}}=1;$$
it follows immediately that $-x-3=0$.
