a \( a^{2 x} \cdot a^{3 x+1} \) b \( 2^{x} \cdot 8^{x+1} \) c \( 3^{2 x} \cdot 27^{x-1} \) d \( \frac{a^{2} b^{3} \times a^{3} b^{4}}{\left(a^{3} b^{3}\right)^{2}} \) g. \( \frac{18^{a} \cdot 15^{a+1}}{18^{a-1} 3 a} \)

a. We start with \[ a^{2x} \cdot a^{3x+1}. \] Using the rule \(\,a^m \cdot a^n = a^{m+n}\), we add the exponents: \[ 2x + (3x+1) = 5x + 1. \] Thus, the expression simplifies to \[ a^{5x+1}. \] --- b. We begin with \[ 2^{x} \cdot 8^{x+1}. \] Since \(8 = 2^3\), we can rewrite the expression as: \[ 2^{x} \cdot (2^3)^{x+1}. \] Using \(\,(2^3)^{x+1} = 2^{3(x+1)} = 2^{3x+3}\), the expression becomes: \[ 2^{x} \cdot 2^{3x+3}. \] Now add the exponents: \[ x + (3x+3) = 4x+3. \] Thus, the simplified form is: \[ 2^{4x+3}. \] --- c. We start with \[ 3^{2x} \cdot 27^{x-1}. \] Writing \(27\) as \(3^3\), we have: \[ 3^{2x} \cdot (3^3)^{x-1} = 3^{2x} \cdot 3^{3(x-1)}. \] Simplify \(3(x-1) = 3x-3\), then add the exponents: \[ 2x + (3x-3) = 5x-3. \] Thus, the expression simplifies to: \[ 3^{5x-3}. \] --- d. The expression is: \[ \frac{a^{2} b^{3} \times a^{3} b^{4}}{\left(a^{3} b^{3}\right)^{2}}. \] First, multiply the factors in the numerator: \[ a^{2} \cdot a^{3} = a^{2+3} = a^{5}, \quad \text{and} \quad b^{3} \cdot b^{4} = b^{3+4} = b^{7}. \] So the numerator becomes: \[ a^{5} b^{7}. \] Now, simplify the denominator. Since \(\left(a^{3} b^{3}\right)^{2} = a^{3 \cdot 2} b^{3 \cdot 2} = a^{6} b^{6}\), we have: \[ \frac{a^{5} b^{7}}{a^{6} b^{6}}. \] Subtracting exponents yields: \[ a^{5-6} b^{7-6} = a^{-1} b^{1} = \frac{b}{a}. \] --- g. We are given: \[ \frac{18^{a} \cdot 15^{a+1}}{18^{a-1} \cdot 3a}. \] First, separate the powers of \(18\): \[ \frac{18^{a}}{18^{a-1}} = 18^{a-(a-1)} = 18^{1} = 18. \] Thus, the expression becomes: \[ 18 \cdot \frac{15^{a+1}}{3a}. \] Divide \(18\) by \(3\): \[ \frac{18}{3} = 6, \] so it simplifies to: \[ 6 \cdot \frac{15^{a+1}}{a}. \] Notice that \(15^{a+1} = 15 \cdot 15^{a}\). Hence, the expression can be written as: \[ 6 \cdot \frac{15 \cdot 15^{a}}{a} = \frac{90 \cdot 15^{a}}{a}. \]