Question

The solution set of the inequality ${\log }_{10}(x^2-16)\le {\log }_{10}(4x-11)$

• A. $(3,5)$
• B. $(4,5)$
• C. $(6,5)$
• D. $noneofthese$

Answer 1

The correct option is B $(4,5)$

$lo{g}_{10}(x^2-16)\le lo{g}_{10}(4x-11)$

$lo{g}_{10}(x^2-16)-lo{g}_{10}(4x-11)\le 0$

$lo{g}_{10}\frac{x^2-16}{4x-11}\le 0$

$log\frac{A}{B}=logA-logB$

$0\le \frac{x^2-16}{4x-11}\le 1$

$x^2-16\le 4x-11$

$x^2-4x+5\le 0$

$(x-5)(x+1)\le 0$

$x\in [-1,5]$ …….$\left(1\right)$

and

$0<x^2-16$

$0<(x-4)(x+4)$

$x\in (-\infty ,-4)\cup (4,\infty )$ ……….$\left(2\right)$

$\left(1\right)$ $cap$ $\left(2\right)$ $=x\in [4,5]$

B is correct.