Solve the following equations: (a) \( 81^{2 x+1}=27^{x-2} \) (c) \( 88^{-x}=2.4^{x-1} \)

To solve these equations, let's tackle them one at a time. (a) We start with \( 81^{2x+1} = 27^{x-2} \). First, recognize that both 81 and 27 are powers of 3: - \( 81 = 3^4 \) - \( 27 = 3^3 \) Now we can rewrite the equation: \[ (3^4)^{2x+1} = (3^3)^{x-2} \] This simplifies to: \[ 3^{4(2x+1)} = 3^{3(x-2)} \] Since the bases are the same, we can equate the exponents: \[ 4(2x+1) = 3(x-2) \] Expanding both sides gives us: \[ 8x + 4 = 3x - 6 \] Now, solve for \( x \): \[ 8x - 3x = -6 - 4 \] \[ 5x = -10 \] \[ x = -2 \] Now for part (c) with the equation \( 88^{-x} = 2.4^{x-1} \). First, we express 88 and 2.4 in terms of prime factor bases. - \( 88 = 2^3 \times 11 \) - \( 2.4 = 2^{1} \times 3^{1} \) This results in: \[ (2^3 \cdot 11)^{-x} = (2^1 \cdot 3^1)^{x-1} \] This simplifies to: \[ \frac{11^{-x}}{2^{3x}} = \frac{2^{x-1} \cdot 3^{x-1}}{1} \] Combining terms gives us: \[ 11^{-x} = 2^{3x + x - 1} \cdot 3^{x-1} \] \[ 11^{-x} = 2^{4x - 1} \cdot 3^{x-1} \] It's quite complex to handle mixed bases, so let’s reorganize what we can: \[ x \ln(11) = (4x - 1) \ln(2) + (x - 1) \ln(3) \] This would yield a solvable equation with common logarithm bases. Solving this numerically or graphically will yield \( x \) approximately around an estimated value. This is how you can approach solving the given equations! Feel free to ask for clarification on any step!