Example Question - solving logarithmic equation
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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 \( x = 8 \).
Solving Logarithmic Equation for x
This question is an equation involving logarithms that we need to solve for x: \( \frac{\log(32)}{\log(x)} + \frac{\log(16)}{\log(x)} = \log(x) \) Firstly, we can combine the terms on the left-hand side using the fact that \( \log(a) + \log(b) = \log(a \cdot b) \). However, since these logarithms have the same denominator, we can combine them as follows: \( \frac{\log(32) + \log(16)}{\log(x)} = \log(x) \) Next, we use the property of logarithms that states \( \log(a) + \log(b) = \log(ab) \) to combine the numerators: \( \frac{\log(32 \cdot 16)}{\log(x)} = \log(x) \) \( \frac{\log(512)}{\log(x)} = \log(x) \) Now we have a single logarithm in the numerator. We can simplify this further by using the property \( \log_b(a) = c \) implies \( b^c = a \): Let's equate the left side to the right side. This gives us an equation of the form: \( \log(512) = \log(x^2) \) Since the logarithms are in the same base and the equation equates one log to another, we can deduce that the arguments must be equal: \( 512 = x^2 \) Taking the square root of both sides gives us: \( x = \sqrt{512} \) \( x = 16 \times \sqrt{2} \) \( x = 16 \times 1.414... \) (since \( \sqrt{2} \) is approximately 1.414) \( x \approx 22.63 \) Therefore, the value of \( x \approx 22.63 \). Please note this value has been rounded to two decimal places.
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