$\log_9(x-1)+\log_4(6-x)=1$

Question

How to solve the equation?

I see solutions $x=2$ and $x=4$. Is there more? Then I try with changing bases:

$$\frac{\log(x-1)}{\log 9}+\frac{\log(6-x)}{\log 4}=1$$

$$\log 4 \log(x-1)+\log 9\log(6-x)=\log 4\log 9$$

I don't know what to do next.

Answer 1

Let $$f(x)=\ln(4)\log(x-1)+\ln(9)\log(6-x)-\ln(4)\ln(9)$$ defined for $1<x<6$.

Then $f$ is smooth on the domain and $$f''(x)=-\frac{\log (4)}{(x-1)^2}-\frac{\log (9)}{(6-x)^2}<0$$ so $f$ is strictly concave and thus has at most $2$ zeros. These zeros are $2$ and $4$ so we are done.

Answer 2

Firstly, $x\in (1,6)$ for the the logarithms to be defined. Let $t=\log_9(x-1)$. Then $ t\in (-\infty, \log_9 5)$ and the equation is equivalent with:

$$9^t+4^{1-t}=5\Leftrightarrow 36^t-5\cdot 4^t=-4$$

Consider the function $f:(-\infty , \log_95)\to \mathbb{R}, f(t)=36^t-5\cdot 4^t$. Then

$$f'(t)=36^t\ln 36-5\cdot 4^t\ln 4$$

$f'(t) \leq 0$ for $t \leq \log_9\frac{5\ln 4}{\ln 36}$ (which means it is decreasing) and $f'(t)\geq 0$ for $t\geq \log_9\frac{5\ln 4}{\ln 36}$ (which means it is increasing). Therefore, it can have at most one solution on each of these intervals, and those are $0$ and $\frac{1}{2}$, which gives $x \in \{2,4\}$ as solution.