\begin{tabular}{|l|l} Solue for \( x \) \\ \hline & \( 5^{x+1} \frac{-3}{15}=2.5^{x} \)\end{tabular}

Let's simplify the equation \( 5^{x+1} \cdot \frac{-3}{15} = 2.5^{x} \). Notice that \( 2.5 \) can be rewritten as \( \frac{5}{2} \), so \( 2.5^{x} = \left(\frac{5}{2}\right)^{x} = \frac{5^x}{2^x} \). Now, substituting this into the equation, we have: \[ 5^{x+1} \cdot \frac{-3}{15} = \frac{5^x}{2^x} \] This simplifies to: \[ \frac{-3 \cdot 5^{x+1}}{15} = \frac{5^x}{2^x} \] We can simplify \( \frac{-3}{15} \) to \( -\frac{1}{5} \), giving us: \[ -\frac{1}{5} \cdot 5^{x+1} = \frac{5^x}{2^x} \] This further simplifies to: \[ -\frac{1}{5} \cdot 5 \cdot 5^x = \frac{5^x}{2^x} \] Which simplifies to: \[ -\frac{5^x}{2} = \frac{5^x}{2^x} \] Now, assuming that \( 5^x \) is not zero (which it isn't for real \( x \)), we’ll divide both sides by \( 5^x \): \[ -\frac{1}{2} = \frac{1}{2^x} \] Next, we can cross-multiply to get: \[ -2 = 2^x \] However, \( 2^x \) cannot equal a negative number, so there are no real values of \( x \) that satisfy the original equation. Thus, the solution to the equation is that there are no solutions in the real number system.