4) \( 16^{x} \cdot 2^{x+1}=\sqrt[5]{4} \)

To solve the equation \( 16^{x} \cdot 2^{x+1}=\sqrt[5]{4} \), we start by expressing everything in terms of base 2. We know that \( 16 = 2^4 \), so we can rewrite \( 16^x \) as \( (2^4)^x = 2^{4x} \). Next, we convert \( \sqrt[5]{4} \) to base 2 as well: since \( 4 = 2^2 \), we have \( \sqrt[5]{4} = (2^2)^{1/5} = 2^{2/5} \). Substituting these back into the equation gives us: \[ 2^{4x} \cdot 2^{x+1} = 2^{2/5} \] By applying the property of exponents \( a^m \cdot a^n = a^{m+n} \), we combine the left side: \[ 2^{4x + (x + 1)} = 2^{2/5} \] \[ 2^{5x + 1} = 2^{2/5} \] Since the bases are the same, we can set the exponents equal to each other: \[ 5x + 1 = \frac{2}{5} \] Now, solving for \( x \): \[ 5x = \frac{2}{5} - 1 = \frac{2}{5} - \frac{5}{5} = \frac{2 - 5}{5} = \frac{-3}{5} \] \[ x = \frac{-3}{5} \cdot \frac{1}{5} = -\frac{3}{25} \] Thus, the solution is: \[ x = -\frac{3}{25} \]