Question

Solve $\log (x+4)+\log (x-4)=\log 3$

• A. $x=-\sqrt{19}$
• B. $x=+\sqrt{19}$
• C. $x=\pm \sqrt{19}$
• D. Cannot be determined

Answer 1

Answer: (B)

$x=+\sqrt{19}$

$\log x+4+\log x-4=\log 3$

$\Rightarrow \log \left\{\right(x+4\left)\right(x-4\left)\right\}=\log 3$

$\Rightarrow \log \{x^2-16\}=\log 3$

$\Rightarrow x^2-16=3\Rightarrow x^2=19$

$\Rightarrow x=\pm \sqrt{19}$

Now, we have to check for $x=+\sqrt{19}$ and $-\sqrt{19}$

If $x=-\sqrt{19};(x-4)<0$

$\Rightarrow x=-\sqrt{19}$ is an invalid solution.

whereas, $x=\sqrt{19}$ is one of the valid solution of the equation given.