7b) Solve using like bases \( \rightarrow 81^{2 x+5}=\left(\frac{1}{3}\right)^{2 x} \)

To solve the equation \( 81^{2x + 5} = \left(\frac{1}{3}\right)^{2x} \), we start by expressing both sides with a base of 3. Since \( 81 = 3^4 \), we can rewrite the left side as \( (3^4)^{2x+5} = 3^{4(2x+5)} = 3^{8x + 20} \). Now for the right side: \(\left(\frac{1}{3}\right)^{2x} = (3^{-1})^{2x} = 3^{-2x}\). So our equation now looks like \( 3^{8x + 20} = 3^{-2x} \). Because both sides have the same base, we can set the exponents equal to each other: \[ 8x + 20 = -2x \] Now, let's solve for \( x \): 1. Add \( 2x \) to both sides: \( 8x + 2x + 20 = 0 \) \( 10x + 20 = 0 \) 2. Subtract 20 from both sides: \( 10x = -20 \) 3. Divide both sides by 10: \( x = -2 \) Therefore, the solution is \( x = -2 \).