Solve for \( x \) in the following equations: \( 2^{x}=8 \) \( 3^{x}=81 \) \( 4^{x}=64 \) f) \( 8^{x+5}=4^{2 x+3} \) \( 2^{x}=\frac{1}{4} \) \( 5^{2 x-1}=125 \) h) \( 8^{x-2}=16^{2 x+1} \) h) \( 2^{\frac{3 x+1}{2}}=\frac{1}{16} \)

To solve each equation for \( x \): 1. \( 2^{x} = 8 \): Since \( 8 = 2^3 \), we have \( 2^{x} = 2^3 \). Therefore, \( x = 3 \). 2. \( 3^{x} = 81 \): Since \( 81 = 3^4 \), we have \( 3^{x} = 3^4 \). Thus, \( x = 4 \). 3. \( 4^{x} = 64 \): Since \( 64 = 4^3 \), we have \( 4^{x} = 4^3 \). Hence, \( x = 3 \). 4. \( 8^{x+5} = 4^{2x+3} \): We know \( 8 = 2^3 \) and \( 4 = 2^2 \), so we can rewrite the equation as: \[ (2^3)^{x+5} = (2^2)^{2x+3} \Rightarrow 2^{3(x+5)} = 2^{2(2x+3)} \] Equate the exponents: \[ 3(x+5) = 4x + 6 \Rightarrow 3x + 15 = 4x + 6 \Rightarrow x = 9 \] 5. \( 2^{x} = \frac{1}{4} \): Rewrite \( \frac{1}{4} \) as \( 2^{-2} \). Thus, we can say: \[ 2^{x} = 2^{-2} \Rightarrow x = -2 \] 6. \( 5^{2x-1} = 125 \): Since \( 125 = 5^3 \), we have: \[ 5^{2x-1} = 5^3 \Rightarrow 2x - 1 = 3 \Rightarrow 2x = 4 \Rightarrow x = 2 \] 7. \( 8^{x-2} = 16^{2x+1} \): Rewrite both sides using powers of 2: \[ (2^3)^{x-2} = (2^4)^{2x+1} \Rightarrow 2^{3(x-2)} = 2^{4(2x+1)} \] Equate the exponents: \[ 3(x-2) = 4(2x+1) \Rightarrow 3x - 6 = 8x + 4 \Rightarrow -5x = 10 \Rightarrow x = -2 \] 8. \( 2^{\frac{3x+1}{2}} = \frac{1}{16} \): Rewrite \( \frac{1}{16} \) as \( 2^{-4} \): \[ 2^{\frac{3x+1}{2}} = 2^{-4} \Rightarrow \frac{3x+1}{2} = -4 \Rightarrow 3x + 1 = -8 \Rightarrow 3x = -9 \Rightarrow x = -3 \] The solutions for \( x \) are: 1. \( x = 3 \) 2. \( x = 4 \) 3. \( x = 3 \) 4. \( x = 9 \) 5. \( x = -2 \) 6. \( x = 2 \) 7. \( x = -2 \) 8. \( x = -3 \)