Question

Number of integral solutins of the equation $lo{g}_{x^3-4}(x^2-1)=lo{g}_{4x-x^2}(x^2-1)$ is

• A. Zero
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

Given: ${\log }_{x^2-4}(x^2-1)={\log }_{4x-x^2}(x^2-1)-\left[1\right]$

where $x^2-4>0,4x-x^2>0$ and $x^2-1>0$

Using ${\log }_{a}b=\frac{{\log }_{a}b}{{\log }_{n}a}$ we can rewrite $\left[1\right]$ as

$\frac{{\log }_{10}(x^2-1)}{{\log }_{10}(x^2-4)}=\frac{{\log }_{10}(x^2-1)}{{\log }_{10}(4x-x^2)}\Rightarrow {\log }_{10}(4x-x^2)={\log }_{10}(x^3-4)\Rightarrow x^3+x^2-4x-4=0\Rightarrow x^2(x+1)-4(x+1)=0\Rightarrow (x^2-4)(x+1)=0\therefore x=-1,2,-2$

However, if $x=-1,{\log }_{x^3-4}(x^2-1)or{\log }_{4x-x^2}(x^2-1)$ is not defined .

For $x=-2;{\log }_{x^3-4}(x^2-1)$ is not defined as $x^3-4<0$

But $x=2$ satisfies the equality $\left[1\right]$ along the constraints mentioned