Solving Logarithmic Equation
The equation shown in the image is:
\[ \frac{\log_2 32 + \log_2 16}{\log_2 x} = \log_2 x \]
To solve for \( x \), we can use logarithmic properties to simplify the left side of the equation and then isolate \( x \). Here are the steps:
1. Simplify the logarithms on the left side of the equation:
\[ \log_2 32 = 5 \] (because \( 2^5 = 32 \))
\[ \log_2 16 = 4 \] (because \( 2^4 = 16 \))
So the equation becomes:
\[ \frac{5 + 4}{\log_2 x} = \log_2 x \]
2. Further simplify the numerator:
\[ \frac{9}{\log_2 x} = \log_2 x \]
3. To isolate \( x \), we can equate the exponents by removing the logarithms. Multiply both sides by \( \log_2 x \) to get rid of the denominator:
\[ 9 = \log_2 x \cdot \log_2 x \]
\[ 9 = (\log_2 x)^2 \]
4. Take the square root of both sides:
\[ \sqrt{9} = \sqrt{(\log_2 x)^2} \]
\[ 3 = |\log_2 x| \]
This gives us two possibilities, because the absolute value of \( \log_2 x \) can be either \( 3 \) or \( -3 \).
Case 1: \( \log_2 x = 3 \)
\[ 2^3 = x \]
\[ x = 8 \]
Case 2: \( \log_2 x = -3 \)
\[ 2^{-3} = x \]
\[ x = \frac{1}{8} \]
However, in the context of the original equation, a negative logarithm would not make sense because it would imply that \( x \) is between 0 and 1, but such \( x \) values would result in a negative denominator in the original equation, which cannot equal a positive \( \log_2 x \) as seen on the right side. Therefore, we discard the second case \( x = \frac{1}{8} \).
The solution to the original equation is \( x = 8 \).
