Solve. \( \left(\frac{1}{5}\right)^{x+4}=25^{x-14} \) \( x=\ldots \)

To solve the equation \( \left(\frac{1}{5}\right)^{x+4}=25^{x-14} \), we first express both sides with the same base. Notice that \( 25 = 5^2 \), so we can rewrite the equation as: \[ \left(\frac{1}{5}\right)^{x+4} = (5^2)^{x-14} \] This simplifies to: \[ \left(5^{-1}\right)^{x+4} = 5^{2(x-14)} \] Applying the power of a power property gives us: \[ 5^{-(x+4)} = 5^{2x - 28} \] Since the bases are the same, we can set the exponents equal to each other: \[ -(x + 4) = 2x - 28 \] Now, solve for \( x \): \[ -x - 4 = 2x - 28 \] \[ -4 + 28 = 2x + x \] \[ 24 = 3x \] \[ x = \frac{24}{3} = 8 \] Thus, the solution is: \[ x = 8 \]