Let's solve the equation [tex]\(3^{x-1} \times 5^{2y-3} = 225\)[/tex].
Step 1: Express 225 in terms of prime factors.
[tex]\[ 225 = 15^2 = (3 \times 5)^2 = 3^2 \times 5^2 \][/tex]
Therefore, we can rewrite the given equation as:
[tex]\[ 3^{x-1} \times 5^{2y-3} = 3^2 \times 5^2 \][/tex]
Step 2: Compare the exponents of the same bases on both sides of the equation.
For the base 3:
[tex]\[ 3^{x-1} = 3^2 \][/tex]
This implies:
[tex]\[ x-1 = 2 \][/tex]
Solving for [tex]\( x \)[/tex], we get:
[tex]\[ x = 3 \][/tex]
For the base 5:
[tex]\[ 5^{2y-3} = 5^2 \][/tex]
This implies:
[tex]\[ 2y-3 = 2 \][/tex]
Solving for [tex]\( y \)[/tex], we get:
[tex]\[ 2y - 3 = 2 \][/tex]
[tex]\[ 2y = 5 \][/tex]
[tex]\[ y = \frac{5}{2} = 2.5 \][/tex]
Therefore, the solution to the equation is:
[tex]\[ x = 3 \quad \text{and} \quad y = 2.5 \][/tex]
