How to find all $m$ for which $\log_3(x - m) + \log_3(x) = \log_3(3x - 4)$ has only one solution over reals?
What I've tried: Domain of the equation is $x > \max(m, \frac{4}{3})$. Using $\log(a) + \log(b) = \log(ab)$ I get $\log_3((x-m)x) = \log_3(3x - 4)$, so $x(x - m) = 3x - 4$. Simplifying yields $x^2 - (m + 3)x + 4 = 0$. Now $\Delta = (m + 3)^2 - 16 = (m - 1)(m + 7)$. For $m = 1$ I get $x = 2$ which satisfies the assumptions. For $m = -7$ I get $x = -2$ which does not satisfy the assumptions. For $m \in (-\infty, -7) \cup (1, \infty)$ there are two solutions $x_1 = \frac{m + 3 + \sqrt{(m + 3)^2 - 16}}{2}$ and $x_2 = \frac{m + 3 - \sqrt{(m + 3)^2 - 16}}{2}$. Now I have to find all $m$ for which $x_1 > \max(m, 4/3)$ and $x_2 \le \max(m, \frac{4}{3})$. Checking it in a straightforward fashion would take quite a long time and wouble be laborious. Is there any faster or better method to solve this? Or maybe am I missing something?
We see that $y=x^2-(m+3)x+4=:f(x)$ is a parabola whose axis of symmetry is $x=\frac{m+3}{2}$.
The key should be to note that $$f\left(\frac 43\right)=\frac{4(4-3m)}{9}\quad\text{and}\quad f(m)=-3m+4$$
For $m\lt -7$, since we have $\frac{m+3}{2}\lt 0$ with $f(0)=4\gt 0$, $f(x)=0$ has no real solutions in $x\gt\frac 43$.
For $1\lt m\lt \frac 43$, we have $$\frac{m+3}{2}\gt 2\gt \frac 43\quad\text{and}\quad f\left(\frac 43\right)=\frac{4(4-3m)}{9}\gt 0$$from which we have that $f(x)=0$ has two real solutions in $x\gt\frac 43$.
For $m\ge \frac 43$, since $f(m)=-3m+4\le 0$, $f(x)=0$ has only one solution in $x\gt m$.
Hence, the answer is $$\color{red}{m=1\quad\text{or}\quad m\ge\frac 43}$$