Question

Precalc.

Answer 1

Domain: x>0.

$(\log_2x)^2+7\log_2x+12=0$

Substitute $t=\log_2x\in\mathbb{R}$. We have:

 $t^2+7t+12=0\\\\a=1\qquad b=7\qquad c=12\\\\\\\Delta=b^2-4ac=7^2-4\cdot1\cdot12=49-48=1\\\\\sqrt{\Delta}=\sqrt{1}=1\\\\\\ t_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-7-1}{2}=\dfrac{-8}{2}=-4\\\\\\ t_2=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-7+1}{2}=\dfrac{-6}{2}=-3$

For t₁ there will be:

$\log_2x=t_1\\\\\log_2x=-4\\\\2^{-4}=x\\\\x=\dfrac{1}{2^4}\\\\\\x=\dfrac{1}{16}=0,0625\\\\\\\boxed{x\approx0,063}$

and for t₂:

$\log_2x=t_2\\\\\log_2x=-3\\\\2^{-3}=x\\\\x=\dfrac{1}{2^3}\\\\\\x=\dfrac{1}{8}\\\\\\\boxed{x=0,125}$

Answer d)