Question

If $x_1$ and $x_2$ satisfy the equation $2^{2x}-{8.2}^x+12=0$ then $x_1+x_2$ equals

• A. 8
• B. ${\log }_{2}6$
• C. ${\log }_{2}12$
• D. $\frac{{\log }_{2}6}{2}$

Answer 1

Answer: (C)

${\log }_{2}12$

Given equation is:

$2^{2x}-{8.2}^x+12=0$

Let $2^x=t$
Hence the equation changes to
$t^2-8t+12=0$
$(t-6)(t-2)=0$
$t=6$ and $t=2$
$\therefore 2^x=6$
$x={\log }_2\left(6\right)$

or
$2^x=2$
$x={\log }_2\left(2\right)=1$
Let $x_1={\log }_{2}6,x_2={\log }_{2}2$

$\therefore x_1+x_2={\log }_2\left(6\right)+{\log }_2\left(2\right)$

$={\log }_2\left(6.2\right)={\log }_2\left(12\right)$