Sure, let's solve the equation step by step.
Given the equation:
[tex]\[
\log_2(x^2) - \log_2(x+2) = 4
\][/tex]
First, we use the properties of logarithms to combine the logs. Specifically, we use the property that [tex]\(\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)\)[/tex]:
[tex]\[
\log_2\left(\frac{x^2}{x+2}\right) = 4
\][/tex]
Next, we convert the logarithmic equation to its exponential form. Recall that if [tex]\(\log_b(A) = C\)[/tex], then [tex]\(b^C = A\)[/tex]:
[tex]\[
2^4 = \frac{x^2}{x+2}
\][/tex]
We know that [tex]\(2^4 = 16\)[/tex], so the equation becomes:
[tex]\[
16 = \frac{x^2}{x+2}
\][/tex]
To clear the fraction, we multiply both sides of the equation by [tex]\(x + 2\)[/tex]:
[tex]\[
16(x + 2) = x^2
\][/tex]
Expanding and rearranging terms to bring all terms to one side of the equation:
[tex]\[
16x + 32 = x^2
\][/tex]
[tex]\[
x^2 - 16x - 32 = 0
\][/tex]
This is a quadratic equation in standard form [tex]\(ax^2 + bx + c = 0\)[/tex], with [tex]\(a = 1\)[/tex], [tex]\(b = -16\)[/tex], and [tex]\(c = -32\)[/tex]. We solve this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[
x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(-32)}}{2(1)}
\][/tex]
[tex]\[
x = \frac{16 \pm \sqrt{256 + 128}}{2}
\][/tex]
[tex]\[
x = \frac{16 \pm \sqrt{384}}{2}
\][/tex]
Since [tex]\(\sqrt{384}\)[/tex] can be simplified to [tex]\(8\sqrt{6}\)[/tex], the equation becomes:
[tex]\[
x = \frac{16 \pm 8\sqrt{6}}{2}
\][/tex]
[tex]\[
x = 8 \pm 4\sqrt{6}
\][/tex]
Thus, the solutions to the equation are:
[tex]\[
x = 8 + 4\sqrt{6} \quad \text{and} \quad x = 8 - 4\sqrt{6}
\][/tex]
To find the largest value of [tex]\(x\)[/tex], we compare the two solutions:
[tex]\[
8 + 4\sqrt{6} \quad \text{and} \quad 8 - 4\sqrt{6}
\][/tex]
Clearly, [tex]\(8 + 4\sqrt{6}\)[/tex] is larger than [tex]\(8 - 4\sqrt{6}\)[/tex].
Therefore, the largest value of [tex]\(x\)[/tex] that satisfies the equation is:
[tex]\[
8 + 4\sqrt{6}
\][/tex]
