Find the number of solutions of a logarithmic equation

Question

$$\log_3(|x-1|)\cdot\log_4(|x-1|)\cdot\log_5(|x-1|) = \log_5(|x-1|) + \log_3(|x-1|)\cdot\log_4(|x-1|)$$ on solving this equation I get this: $$|x-1|/5 = 4\log_{|x-1|}3$$ What should I do next??

Answer 1

Let $t=\log_3(|x-1|)$ then $\log_4(|x-1|)=\frac{t}{\log_3(4)}$, $\log_5(|x-1|)=\frac{t}{\log_3(5)}$, and the equation becomes $$\frac{t^3}{\log_3(4)\cdot \log_3(5)}=\frac{t}{\log_3(5)}+\frac{t^2}{\log_3(4)}.$$ which implies that $t=0$ (and therefore $x=2$ or $x=0$) or $t$ is a solution of this quadratic equation $$t^2-\log_3(5)t-\log_3(4)=0.$$ Can you take it from here?

Answer 2

Hint: Write your equation in the form $$\frac{\ln|x-1|}{\ln(3)}\frac{\ln|x-1|}{\ln(4)}\frac{\ln|x-1|}{\ln(5)}=\frac{\ln|x-1|}{\ln(5)}+\frac{\ln|x-1|}{\ln(3)}\frac{\ln|x-1|}{\ln(4)}$$

Answer 3

The equation that you have arrived at is wrong; it does not have the same solutions.

Because $\log_ab=\frac{\log b}{\log a}$ the equation can be written as $$\frac{\log|x-1|}{\log 3}\frac{\log|x-1|}{\log 4}\frac{\log|x-1|}{\log 5}=\frac{\log|x-1|}{\log 5}+\frac{\log|x-1|}{\log 3}\frac{\log|x-1|}{\log 4},$$ and setting $y:=\log|x-1|$ this simplifies to $$\frac{1}{\log3\log4\log5}y^3-\frac{1}{\log3\log4}y^2-\frac{1}{\log5}y=0.$$ One obvious solution is $y=0$, corresponding to $x=0$ and $x=2$. If $y\neq0$ then we can divide by $y$ to get $$\frac{1}{\log3\log4\log5}y^2-\frac{1}{\log3\log4}y-\frac{1}{\log5}=0,$$ which is a quadratic equation in $y$, which I'm sure you can solve.