4. \( 3^{7 x}=9.2^{7 x} \)

To solve the equation \( 3^{7x} = 9 \cdot 2^{7x} \), we start by rewriting \( 9 \) as \( 3^2 \). This gives us: \[ 3^{7x} = 3^2 \cdot 2^{7x} \] Next, we can express the left side in terms of \( 3^{7x} \): \[ 3^{7x} = 3^2 \cdot 2^{7x} \] Now, we can divide both sides by \( 3^{7x} \): \[ 1 = \frac{3^2}{3^{7x}} \cdot 2^{7x} \] This simplifies to: \[ 1 = 3^{2 - 7x} \cdot 2^{7x} \] Taking the logarithm of both sides (we can use any logarithm, here we will use natural logarithm \( \ln \)) gives: \[ \ln(1) = \ln(3^{2 - 7x}) + \ln(2^{7x}) \] Since \( \ln(1) = 0 \), we have: \[ 0 = (2 - 7x) \ln(3) + (7x) \ln(2) \] Rearranging terms leads to: \[ 7x \ln(2) = - (2 - 7x) \ln(3) \] Distributing the \( - \ln(3) \): \[ 7x \ln(2) = -2 \ln(3) + 7x \ln(3) \] Now, we bring all \( x \) terms to one side: \[ 7x \ln(2) - 7x \ln(3) = -2 \ln(3) \] Factoring out \( 7x \): \[ 7x (\ln(2) - \ln(3)) = -2 \ln(3) \] Now solving for \( x \): \[ x = \frac{-2 \ln(3)}{7 (\ln(2) - \ln(3))} \] This is the solution to the equation.