Rewrite $5^{12x-17}=125$ as a logarithm. Then apply the change of base formula to solve for x using the common log. Round to the nearest thousandth. I attempted the question in the title:

Rewrite $5^{12x-17}=125$ as a logarithm. Then apply the change of base formula to solve for x using the common log. Round to the nearest thousandth.

I arrived at $x=\frac{14}{12}$ whereas my textbook says the solution is actually this:

My working:
$$5^{12x-17}=125$$
$$\log_5(125)=12x-17$$
$$\frac{\ln(125)}{\ln(5)}=12x-17$$
$$3=12x-17$$
$$12x=14$$
$$x=\frac{14}{12}$$
Where did I go wrong and how can I arrive at $\frac{5}{3}$?

Alternatively, from $5^{12x-17} = 5^3$ you can directly conclude that $12x-17 = 3 \implies x = \frac{5}{3}$ as the exponential function $a^x$ is injective (one-to-one). It is injective as its inverse exists, being the function $\frac{\ln x}{\ln a}$, which you can find by making $x$ the subject in $y = a^x$.
Note that this does not hold in general. For instance, $x^2 = 4 \implies x = 2, -2$ as when drawing the line $y = 4$, it intersects the curve at two points, hence two solutions. Since $12x-17$ can be any real number, you need this property of $1$ y-value $\implies$ $1$ x-value to hold for all real $x$.
