Question

solve: $(log_{2}(x))^{2} - 9log_{2}(x) = -18$.

a. $x = 8$ and $x = 64$

b. $x = 5.9$

c. $x = \frac{1}{8}$ and $x = \frac{1}{64}$

d. $x = 3$ and $x = 6$

Explanation:

Step1: Substitute variable

Let $y = \log_2(x)$. The equation becomes:
$$y^2 - 9y = -18$$

Step2: Rearrange to standard quadratic

Rewrite into $ay^2+by+c=0$ form:
$$y^2 - 9y + 18 = 0$$

Step3: Factor the quadratic

Factor the polynomial:
$$(y - 3)(y - 6) = 0$$

Step4: Solve for y

Set each factor equal to 0:
$y - 3 = 0 \implies y=3$; $y - 6 = 0 \implies y=6$

Step5: Substitute back to find x

For $y=3$: $\log_2(x)=3 \implies x=2^3=8$
For $y=6$: $\log_2(x)=6 \implies x=2^6=64$

Answer:

A. $x = 8$ and $x = 64$