Questions: 2 log4 x - 3 log4 x = log4 48 - log4 3

Solution Steps

To solve the equation \(2 \log_{4}^{2} x - 3 \log_{4} x = \log_{4} 48 - \log_{4} 3\), we can use properties of logarithms. First, simplify the right side using the quotient rule of logarithms. Then, let \(y = \log_{4} x\) to transform the equation into a quadratic form. Solve the quadratic equation for \(y\), and then solve for \(x\) by converting back from the logarithmic form.

The original equation is: \[ 2 \log_{4}^{2} x - 3 \log_{4} x = \log_{4} 48 - \log_{4} 3 \]

Using the quotient rule of logarithms, simplify the right side: \[ \log_{4} 48 - \log_{4} 3 = \log_{4} \left(\frac{48}{3}\right) = \log_{4} 16 \]

Since \(16 = 4^2\), we have: \[ \log_{4} 16 = 2 \]

Let \( y = \log_{4} x \). Substitute into the equation: \[ 2y^2 - 3y = 2 \]

Rearrange to form a quadratic equation: \[ 2y^2 - 3y - 2 = 0 \]

Solve the quadratic equation \(2y^2 - 3y - 2 = 0\) using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 2\), \(b = -3\), and \(c = -2\).

The solutions are: \[ y = -\frac{1}{2} \quad \text{and} \quad y = 2 \]

Convert back to \(x\) using \(y = \log_{4} x\):

• For \(y = -\frac{1}{2}\): \[ \log_{4} x = -\frac{1}{2} \] \[ x = 4^{-\frac{1}{2}} = \frac{1}{\sqrt{4}} = \frac{1}{2} \]

• For \(y = 2\): \[ \log_{4} x = 2 \] \[ x = 4^2 = 16 \]

Final Answer