To solve the equation [tex]\(\log_{16}(x) = (\log_4(x))^2\)[/tex] for [tex]\(x > 1\)[/tex], we follow these steps:
1. Rewrite the logarithms using a common base:
We know that [tex]\(16\)[/tex] can be written as [tex]\(4^2\)[/tex]. Therefore, we can express [tex]\(\log_{16}(x)\)[/tex] in terms of [tex]\(\log_4(x)\)[/tex]:
[tex]\[
\log_{16}(x) = \frac{\log_4(x)}{\log_4(16)}
\][/tex]
Since [tex]\(16 = 4^2\)[/tex], we have [tex]\(\log_4(16) = 2\)[/tex]:
[tex]\[
\log_{16}(x) = \frac{\log_4(x)}{2}
\][/tex]
2. Substitute this expression into the original equation:
Substitute [tex]\(\log_{16}(x) = \frac{\log_4(x)}{2}\)[/tex] into the equation [tex]\(\log_{16}(x) = (\log_4(x))^2\)[/tex]:
[tex]\[
\frac{\log_4(x)}{2} = (\log_4(x))^2
\][/tex]
3. Set up the equation:
Let [tex]\(y = \log_4(x)\)[/tex]. Then the equation becomes:
[tex]\[
\frac{y}{2} = y^2
\][/tex]
4. Solve the quadratic equation:
Multiply both sides by 2 to clear the fraction:
[tex]\[
y = 2y^2
\][/tex]
Rearrange the equation:
[tex]\[
2y^2 - y = 0
\][/tex]
Factor out the common term [tex]\(y\)[/tex]:
[tex]\[
y(2y - 1) = 0
\][/tex]
5. Find the solutions for [tex]\(y\)[/tex]:
Set each factor equal to zero:
[tex]\[
y = 0 \quad \text{or} \quad 2y - 1 = 0
\][/tex]
Solve each equation:
[tex]\[
y = 0 \quad \text{or} \quad y = \frac{1}{2}
\][/tex]
6. Back-substitute [tex]\(\log_4(x) = y\)[/tex]:
Recall that [tex]\(y = \log_4(x)\)[/tex]:
[tex]\[
\log_4(x) = 0 \quad \text{or} \quad \log_4(x) = \frac{1}{2}
\][/tex]
7. Solve for [tex]\(x\)[/tex]:
For [tex]\(\log_4(x) = 0\)[/tex]:
[tex]\[
x = 4^0 = 1
\][/tex]
For [tex]\(\log_4(x) = \frac{1}{2}\)[/tex]:
[tex]\[
x = 4^{1/2} = \sqrt{4} = 2
\][/tex]
8. Consider the condition [tex]\(x > 1\)[/tex]:
Since the problem specifies [tex]\(x > 1\)[/tex], we discard [tex]\(x = 1\)[/tex]. Thus, the valid solution is:
[tex]\[
x = 2
\][/tex]
Therefore, the solution to the equation [tex]\(\log_{16}(x) = (\log_4(x))^2\)[/tex] for [tex]\(x > 1\)[/tex] is:
[tex]\[
x = 2
\][/tex]
