Question
Real Tutor Solution
Tutor-Verified Answer
Answer
The solutions are \( y = 8 \) and \( y = \frac{1}{8} \).
Solution
To solve the equation:
\[
\log_{2} y = \frac{9}{\log_{2} y}
\]
Let's denote \( x = \log_{2} y \). Substituting this into the equation gives:
\[
x = \frac{9}{x}
\]
Multiply both sides by \( x \) to eliminate the fraction:
\[
x^2 = 9
\]
Solving for \( x \):
\[
x = \sqrt{9} \quad \text{or} \quad x = -\sqrt{9}
\]
\[
x = 3 \quad \text{or} \quad x = -3
\]
Now, recall that \( x = \log_{2} y \), so we can solve for \( y \) in each case:
1. **Case 1: \( x = 3 \)**
\[
\log_{2} y = 3 \implies y = 2^3 = 8
\]
2. **Case 2: \( x = -3 \)**
\[
\log_{2} y = -3 \implies y = 2^{-3} = \frac{1}{8}
\]
**Verification:**
- For \( y = 8 \):
\[
\log_{2} 8 = 3 \quad \text{and} \quad \frac{9}{\log_{2} 8} = \frac{9}{3} = 3
\]
So, \( 3 = 3 \) ✔️
- For \( y = \frac{1}{8} \):
\[
\log_{2} \left(\frac{1}{8}\right) = -3 \quad \text{and} \quad \frac{9}{\log_{2} \left(\frac{1}{8}\right)} = \frac{9}{-3} = -3
\]
So, \( -3 = -3 \) ✔️
**Conclusion:**
The solutions to the equation are \( y = 8 \) and \( y = \frac{1}{8} \).
**Answer:** Both solutions are y = 8 and y = 1⁄8.
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Bonus Knowledge
To solve the equation \( \log _{2} y=\frac{9}{\log _{2} y} \), we can start by letting \( x = \log_{2} y \). The equation then becomes \( x = \frac{9}{x} \). Multiplying both sides by \( x \) gives \( x^2 = 9 \), leading to \( x = 3 \) or \( x = -3 \). This means \( \log_{2} y = 3 \) implies \( y = 2^3 = 8 \), while \( \log_{2} y = -3 \) implies \( y = 2^{-3} = \frac{1}{8} \). Therefore, the two solutions for \( y \) are \( 8 \) and \( \frac{1}{8} \). Feel free to ask for more details about these solutions or further assistance with logarithmic equations!
