Question - Solving Logarithmic Equation
Solution:
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 $$ x = 8 $$.
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