d. \( 5^{2 x+4}-25^{x-1}=78120 \)

To solve the equation \( 5^{2x+4} - 25^{x-1} = 78120 \), we can start by rewriting \( 25^{x-1} \) in terms of base 5. Since \( 25 = 5^2 \), we have: \[ 25^{x-1} = (5^2)^{x-1} = 5^{2(x-1)} = 5^{2x-2} \] Now substituting this back into the equation gives us: \[ 5^{2x+4} - 5^{2x-2} = 78120 \] Next, we can factor out \( 5^{2x-2} \): \[ 5^{2x-2}(5^6 - 1) = 78120 \] Since \( 5^6 = 15625 \), we have: \[ 5^{2x-2}(15625 - 1) = 78120 \] This simplifies to: \[ 5^{2x-2} \cdot 15624 = 78120 \] Now, divide both sides by 15624: \[ 5^{2x-2} = \frac{78120}{15624} \] Calculating the right side gives: \[ 5^{2x-2} = 5 \] Now, we can rewrite \( 5 \) as \( 5^1 \): \[ 5^{2x-2} = 5^1 \] Since the bases are the same, we can set the exponents equal to each other: \[ 2x - 2 = 1 \] Now, solve for \( x \): \[ 2x = 3 \quad \Rightarrow \quad x = \frac{3}{2} \] Thus, the solution to the equation is: \[ \boxed{\frac{3}{2}} \]