The question is :
$f(x) = \log_3 x, g(x) = \log_3{(x+3)+4}$
There are two intersections $\mathrm A$ and $\mathrm B$ made by $y=-{4\over 3}x+4$
$\overline {\mathrm {AB}} = ?$
It was obvious by the graph that both log graphs are translated with the same scale of the line, so the answer is 5 by applying pythagorean theorem. But I can't figure out how to prove this with equations only.
More generally, I wanted to show if there is a line $y={b \over a}x + c$, and the only intersections $\mathrm P$, $\mathrm Q$ are each made by $f(x)$, $f(x-a)+b$, there has to be a relation of $q = p+a$ where $p$, $q$ are $x$-component of $\mathrm P$ and $\mathrm Q$.
I know it seems so obvious that it may sound silly, but is there a way to rigorously prove it with a algebra?
Only idea I could think of was setting $\mathrm P(p, f(p))$ , $\mathrm Q(q, f(q-a)+b)$ and showing that the slope of two points is $b \over a$. But that was impossible because there was no way I can proceed by using equations.
Since $P$ is on the line $y=\frac{b}{a}x+c$, $p$ satisfies $f(p)=\frac{b}{a}p+c$. You can verify that the point $(p+a,f(p)+b)$ is also on the line by verifying $f(p)+b=\frac{b}{a}(p+a)+c$. You can also verify that the point is on the curve $y=f(x-a)+b$, So this point is the intersection of the line and the curve, which must be $Q$.