To solve the equation

$\left[\log_2(x)\right]^2 + 7 \log_2(x^2) + 48 = 0$

we'll proceed with the following steps.

Step 1: Simplify the Logarithmic Terms

The term $\log_2(x^2)$ can be simplified using the power rule of logarithms, which states $\log_b(a^c) = c \cdot \log_b(a)$.

$\log_2(x^2) = 2 \cdot \log_2(x)$

So the equation becomes:

$\left[\log_2(x)\right]^2 + 7 \cdot 2 \cdot \log_2(x) + 48 = 0$

or

$\left[\log_2(x)\right]^2 + 14 \log_2(x) + 48 = 0$

Step 2: Substitute a Variable

Let $y = \log_2(x)$. Then $\left[\log_2(x)\right]^2 = y^2$ and $\log_2(x) = y$.

Substituting these into the equation gives:

$y^2 + 14y + 48 = 0$

Step 3: Solve the Quadratic Equation

The equation $y^2 + 14y + 48 = 0$ is a standard quadratic equation in $y$. We can solve it using the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For our equation, $a = 1$, $b = 14$, and $c = 48$. Plugging these values into the formula:

$y = \frac{-14 \pm \sqrt{14^2 - 4 \cdot 1 \cdot 48}}{2 \cdot 1}$

$y = \frac{-14 \pm \sqrt{196 - 192}}{2}$

$y = \frac{-14 \pm \sqrt{4}}{2}$

$y = \frac{-14 \pm 2}{2}$

This gives two solutions for $y$:

1. $y = \frac{-14 + 2}{2} = \frac{-12}{2} = -6$
2. $y = \frac{-14 - 2}{2} = \frac{-16}{2} = -8$

Step 4: Substitute Back for $x$

Recall that $y = \log_2(x)$. So we have two equations:

1. $\log_2(x) = -6$
2. $\log_2(x) = -8$

To solve for $x$, rewrite each equation in exponential form:

1. $x = 2^{-6} = \frac{1}{64}$
2. $x = 2^{-8} = \frac{1}{256}$

Final Answer

The solutions are:

$x = \frac{1}{64} \quad \text{and} \quad x = \frac{1}{256}$

Would you like a more detailed explanation, or do you have any questions?

Related Questions

1. How would the solution change if the base of the logarithm were different?
2. Can logarithmic equations always be transformed into quadratic equations?
3. What if the constant term was negative; how would it affect the solution?
4. Are there alternate methods to solve logarithmic equations without substitution?
5. How does changing the argument of the logarithm affect the equation?

Tip

Always check for domain restrictions when working with logarithmic equations, as the argument of a logarithm must be positive.