Question

The number of solutions of the equation ${\log }_4\left(x-1\right)={\log }_2\left(x-3\right)$ is

Answer 1

Enumerate the number of possible solutions

Given equation is ${\log }_4\left(x-1\right)={\log }_2\left(x-3\right)$
$$\begin{gathered} \begin{array}{ccc}\Rightarrow & 4^{{\log }_4\left(x-1\right)}=4^{{\log }_2\left(x-3\right)}& \\ \Rightarrow & x-1={\left(2^2\right)}^{{\log }_2\left(x-3\right)}& \left[\because a^{{\log }_a\left(x\right)}=x\right]\\ \Rightarrow & x-1=2^{2{\log }_2\left(x-3\right)}& \left[\because {\left(a^m\right)}^n=a^{m\times n}\right]\\ \Rightarrow & x-1=2^{{\log }_2{\left(x-3\right)}^2}& \left[\because x\log a=\log \left(a^x\right)\right]\\ \Rightarrow & x-1={\left(x-3\right)}^2& \\ \Rightarrow & x-1=x^2+9-6x& \\ \Rightarrow & x^2+10-7x=0& \\ \Rightarrow & x^2-2x-5x+10=0& \\ \Rightarrow & x\left(x-2\right)-5\left(x-2\right)=0& \\ \Rightarrow & \left(x-2\right)\left(x-5\right)=0& \end{array} \\ \begin{array}{ccc}\Rightarrow & \left(x-2\right)=0& \left(x-5\right)=0\\ \Rightarrow & x=2& x=5\end{array} \end{gathered}$$

When $x=2$,
$\begin{array}{rcl}{\log }_2\left(x-3\right)& =& {\log }_2\left(2-3\right)\\ & =& {\log }_2\left(-1\right)\end{array}$

But $\log \left(x\right)$ is not defined for $x\le 0$

So, only $x=5$ is a valid solution.

Hence, the number of solutions to the equation ${\log }_4\left(x-1\right)={\log }_2\left(x-3\right)$ is $1$.