How does $49\cdot\left(\frac{1}{7}\right)^{x}-\left(\frac{1}{7}\right)^x$ become $\left(\frac{1}{7}\right)^{x}\left(49-1\right)$? Rewrite $\left(\frac{1}{7}\right)^{x-2}-\left(\frac{1}{7}\right)^x$ as $A\cdot\left(\frac{1}{7}\right)^x$ is a question I got.
Now the answer to A is 48, but I don't really get how they got to the answer:
$$\left(\frac{1}{7}\right)^{x-2}-\left(\frac{1}{7}\right)^x = \frac{\left(\frac{1}{7}\right)^{x}}{\left(\frac{1}{7}\right)^{2}}-\left(\frac{1}{7}\right)^x$$
$$ = 49\cdot\left(\frac{1}{7}\right)^{x}-\left(\frac{1}{7}\right)^x$$
so far this makes sense, multiplying by the reciprocal of $\frac{1}{\left(\frac{1}{7}\right)^2}$ ends up in $\left(\frac{7}{1}\right)^2$ I suppose, so 49. But the next step does not make any sense to me:
$$ = \left(\frac{1}{7}\right)^{x}\left(49-1\right)$$
How do we end up with 49 minus 1? Or do I have to treat this expression the same as $10x - x$ which is the same as $10x - 1x$, so we end up with $9x$. And in this specific case the $x$'s are the fractions to the $x$ power?

Note that, for all $a$ and $b$, due to the distributive property, you have
$$(a-1)b = ab - b \tag{1}\label{eq1A}$$
Thus, with $a = 49$ and $b = \left(\frac{1}{7}\right)^x$, you get what you're asking about, i.e.,
$$49\left(\frac{1}{7}\right)^x - \left(\frac{1}{7}\right)^x = (49 - 1)\left(\frac{1}{7}\right)^x \tag{2}\label{eq2A}$$
