Question

$5^x+(2\sqrt{3}{)}^{2x}-169\le 0$ is true in the interval :

• A. $(-\infty ,2)$
• B. $(0,2)$
• C. $(2,\infty )$
• D. $(0,4)$

Answer 1

The correct option is A $(-\infty ,2)$

$5^x+(2\sqrt{3}{)}^{2x}-169$

$=5^x+\left(\right(2\sqrt{3}{)}^2{)}^x-169$

$=5^x+{12}^x-169$

When $x=2,5^2+{12}^2-169=0$

$\therefore$ When $x\le 2,5^x+{12}^x-169\le 0$

$\therefore 5^x+{12}^x-169\le 0$, for $x\in (-\infty ,2]$