Example Question - isolating variable x
Here are examples of questions we've helped users solve.
Solving a Linear Equation
Задача, которую вы хотели бы решить, - это номер 3: 1,5x + 6,3 = x + 18. Давайте решим это уравнение пошагово: Сначала перенесём 'x' с правой стороны уравнения на левую, вычитая 'x' из обеих частей: 1,5x - x + 6,3 = 18, Теперь упростим левую сторону, вычитая коэффициенты при 'x': 0,5x + 6,3 = 18. Теперь вычтем 6,3 из обеих частей уравнения, чтобы изолировать переменную 'x': 0,5x = 18 - 6,3, 0,5x = 11,7. Теперь, чтобы найти значение 'x', разделим обе части уравнения на 0,5: x = 11,7 / 0,5, x = 23,4. Ответ: x = 23,4.
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 \( x = 8 \).
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