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

The correct option is B $x=+\sqrt{19}$

We know that

$\log A+\log B=\log AB$

So, $\log (x+4)+\log (x-4)=\log 3$

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

$\Rightarrow (x+4)(x-4)=3$

$\Rightarrow x^2-16=3$

$\Rightarrow x^3=19$

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

Log of a negative number does not exist, hence $x=+\sqrt{19}$