27. \( 2^{3 x}=4 \cdot 2^{5 x} \) 29. \( (1+x) 2^{-x}-5 \cdot 2^{-x}=0 \)

Let's solve each of the equations step by step. --- **Problem 27.** \[ 2^{3x} = 4 \cdot 2^{5x} \] **Step 1:** Express the number 4 as a power of 2: \[ 4 = 2^2 \] So the equation becomes: \[ 2^{3x} = 2^2 \cdot 2^{5x} \] **Step 2:** Use the property of exponents \( a^m \cdot a^n = a^{m+n} \): \[ 2^{3x} = 2^{2 + 5x} \] **Step 3:** Since the bases are the same, set the exponents equal to each other: \[ 3x = 2 + 5x \] **Step 4:** Solve for \( x \): \[ \begin{align*} 3x &= 2 + 5x \\ 3x - 5x &= 2 \\ -2x &= 2 \\ x &= -1 \end{align*} \] **Answer:** \( x = -1 \) --- **Problem 29.** \[ (1 + x) \cdot 2^{-x} - 5 \cdot 2^{-x} = 0 \] **Step 1:** Factor out \( 2^{-x} \) from both terms: \[ 2^{-x} \cdot (1 + x - 5) = 0 \] **Step 2:** Simplify inside the parentheses: \[ 2^{-x} \cdot (x - 4) = 0 \] **Step 3:** Set each factor equal to zero: 1. \( 2^{-x} = 0 \) However, \( 2^{-x} \) is never zero for any real number \( x \), so this gives no solution. 2. \( x - 4 = 0 \) Solve for \( x \): \[ x = 4 \] **Answer:** \( x = 4 \) --- **Summary of Solutions:** - **Problem 27:** \( x = -1 \) - **Problem 29:** \( x = 4 \)