Simplify. (9^(2y)×3^(y))/(27^(-(y)/(3)))

Express with same base: We start by expressing all the terms in the expression with the same base, which is 3, because 9 and 27 are both powers of 3. 9 = 3^2 and 27 = 3^3Rewrite using base of 3: Rewrite the expression using the base of 3. (9^(2y)×3^(y))/(27^(-(y)/(3))) = ((3^2)^(2y)×3^(y))/((3^3)^(-(y)/(3)))Apply power rules: Apply the power of a power rule (a^(m))^n = a^(m*n) and the power of a product rule a^m * a^n = a^(m+n). ((3^2)^(2y)×3^(y))/((3^3)^(-(y)/(3))) = (3^(2*2y)×3^(y))/(3^(3*(-(y)/(3)))) = (3^(4y)×3^(y))/(3^(-y))Combine exponents: Combine the exponents in the numerator using the power of a product rule. 3^(4y)×3^(y) = 3^(4y+y) = 3^(5y)Apply quotient rule: Now we have: (3^(5y))/(3^(-y)) Apply the quotient of powers rule a^m / a^n = a^(m-n). 3^(5y) / 3^(-y) = 3^(5y - (-y)) = 3^(5y + y) = 3^(6y)Final simplified form: The expression is now simplified to: 3^(6y) This is the final simplified form of the given expression.

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Simplify.
(9^(2y)×3^(y))/(27^(-(y)/(3)))

Simplify. $\newline$ $\frac{9^{2 y} \times 3^{y}}{27^{-\frac{y}{3}}}$

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Math Problems

Algebra 1

Evaluate integers raised to rational exponents

Full solution

Q. Simplify.

\newline

\frac{9^{2 y} \times 3^{y}}{27^{-\frac{y}{3}}}

Express with same base: We start by expressing all the terms in the expression with the same base, which is $3$ , because $9$ and $27$ are both powers of $3$ . $\newline$ $9 = 3^2$ and $27 = 3^3$
Rewrite using base of $3$ : Rewrite the expression using the base of $3$ . $\newline$ $(9^{2y}\times3^{y})/(27^{-\frac{y}{3}})$ $\newline$ = $((3^2)^{2y}\times3^{y})/((3^3)^{-\frac{y}{3}})$
Apply power rules: Apply the power of a power rule $(a^{m})^{n} = a^{m*n}$ and the power of a product rule $a^{m} \times a^{n} = a^{m+n}$ . $((3^{2})^{(2y)}\times3^{y})/((3^{3})^{-(y)/(3)})$ = $(3^{2*2y}\times3^{y})/(3^{3*(-(y)/(3))})$ = $(3^{4y}\times3^{y})/(3^{-y})$
Combine exponents: Combine the exponents in the numerator using the power of a product rule. $3^{4y}\times3^{y} = 3^{4y+y} = 3^{5y}$
Apply quotient rule: Now we have: $\newline$ $\frac{3^{5y}}{3^{-y}}$ $\newline$ Apply the quotient of powers rule $a^m / a^n = a^{(m-n)}$ . $\newline$ $\frac{3^{5y}}{3^{-y}} = 3^{(5y - (-y))} = 3^{(5y + y)} = 3^{6y}$
Final simplified form: The expression is now simplified to: $3^{6y}$ This is the final simplified form of the given expression.