Question

$\text{The value of}x\text{for which}{\log }_3(x+4)+{\log }_3(x-1)=1\text{is}$

• A. $2$
• B. $4$
• C. $1$

Answer 1

The correct option is C $1$

Given: ${\log }_3(x+4)+{\log }_3(x-1)=1$
Now, for logarithm to be defined:
$x+4>0\&x-1>0\Rightarrow x>-4\&x>1\Rightarrow x>1\cdots \left(A\right)$

Also, Using the property:
${\log }_{a}p+{\log }_{a}q={\log }_a\left(pq\right)$

Thus, ${\log }_3(x+4)+{\log }_3(x-1)=1$
$\Rightarrow {\log }_3\left\{\right(x+4\left)\right(x-1\left)\right\}=1$
Now, Using the proeprty:
${\log }_{a}a=1$
Thus, ${\log }_3\left\{\right(x+4\left)\right(x-1\left)\right\}=1={\log }_{3}3$
$\Rightarrow (x+4)(x-1)=3\Rightarrow x^2+3x-4=3\Rightarrow x^2+3x-7=0$
Using the quadratic formulae, we get:
$x=\frac{-3\pm \sqrt{9+4\times 7}}{2}\Rightarrow x=\frac{-3\pm \sqrt{37}}{2}\Rightarrow x=\frac{-3+\sqrt{37}}{2}\&\frac{-3-\sqrt{37}}{2}$
Now, $\frac{-3+\sqrt{37}}{2}\in (1,\infty )$
Thus, there is only one value of $x$ that satsisfies the given equation.