Question

The value of $x$ if ${\log }_{10}{x}^2-7x-6=1-{\log }_{10}5$

• A. $-4$
• B. $-1$
• C. $2$
• D. $8$

Answer 1

Answer: (B), (D)

The correct options are
B $-1$
D $8$

Given:
${\log }_{10}{x}^2-7x-6=1-{\log }_{10}5$
For the $\log$ to be defined,
$x^2-7x-6>0$
So, the roots of equation
$x^2-7x-6=0\Rightarrow x=\frac{7\pm \sqrt{73}}{2}$
So,
$x\in (-\infty ,\frac{7-\sqrt{73}}{2})\cup (\frac{7+\sqrt{73}}{2},\infty )\cdots \left(1\right)$

Now,
${\log }_{10}{x}^2-7x-6=1-{\log }_{10}5\Rightarrow {\log }_{10}{x}^2-7x-6+{\log }_{10}5=1\Rightarrow {\log }_{10}5(x^2-7x-6)=1\Rightarrow 5(x^2-7x-6)=10$
$\Rightarrow x^2-7x-6=2$
$\Rightarrow x^2-7x-8=0$
$\Rightarrow x^2-8x+x-8=0$
$\Rightarrow (x+1)(x-8)=0$
​​​​​​​​​​​​​​$\Rightarrow x=-1,8$
From equation $\left(1\right)$, both values are accepted
$\therefore x=-1,8$