Question

Precalc.

Solve the equation for x, accurate to three decimal places: (log2x)^2 + 7log2x + 12 = 0

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\\\\√(\Delta)=√(1)=1\\\\\\ t_1=(-b-√(\Delta))/(2a)=(-7-1)/(2)=(-8)/(2)=-4\\\\\\ t_2=(-b+√(\Delta))/(2a)=(-7+1)/(2)=(-6)/(2)=-3`

For t₁ there will be:


`\log_2x=t_1\\\\\log_2x=-4\\\\2^(-4)=x\\\\x=(1)/(2^4)\\\\\\x=(1)/(16)=0,0625\\\\\\\boxed{x\approx0,063}`

and for t₂:


`\log_2x=t_2\\\\\log_2x=-3\\\\2^(-3)=x\\\\x=(1)/(2^3)\\\\\\x=(1)/(8)\\\\\\\boxed{x=0,125}`

Answer d)