a) Solve for \( x \) \( 4 \cdot 4^{x}=16 \) \( 12,8^{x} \cdot 2=128 \) \( 14,4^{x}=\frac{1}{16} \)

To solve for \( x \), let's tackle each equation one by one. 1. For \( 4 \cdot 4^{x} = 16 \): We can rewrite 16 as \( 4^2 \), so: \[ 4^{1 + x} = 4^2 \] This gives us \( 1 + x = 2 \), leading to: \[ x = 1 \] 2. For \( 12 \cdot 8^{x} \cdot 2 = 128 \): Rewrite \( 128 \) as \( 2^7 \) and \( 8 \) as \( 2^3 \): \[ 12 \cdot 2 \cdot (2^3)^{x} = 2^7 \] Simplifying gives: \[ 24 \cdot 2^{3x} = 2^7 \] We have \( 24 = 2^3 \cdot 3 \), so: \[ 2^3 \cdot 3 \cdot 2^{3x} = 2^7 \] This simplifies to: \[ 2^{3 + 3x} \cdot 3 = 2^7 \] Dividing both sides by \( 3 \) won't eliminate it entirely, so we compare powers: \[ 3 + 3x = 7 \implies 3x = 4 \implies x = \frac{4}{3} \] 3. For \( 14 \cdot 4^{x} = \frac{1}{16} \): Rewrite \( \frac{1}{16} \) as \( 4^{-2} \): \[ 14 \cdot 4^{x} = 4^{-2} \] Dividing both sides by 14, we have: \[ 4^{x} = \frac{4^{-2}}{14} \] Since \( 4^{-2} = \frac{1}{16} \), we get: \[ 4^{x} = \frac{1}{14 \cdot 16} \] To find the exponent \( x \), we can take logarithms or notice that: \[ x = -2 - \log_4{14} \] Thus, the solutions are: 1. \( x = 1 \) 2. \( x = \frac{4}{3} \) 3. \( x ≈ -2 - \log_4{14} \)