Which expression is equivalent to ((3^(2))^(-6))/(3^(4))? (1)/(3^(17)) 3^(16) (1)/(3^(18)) (1)/(3^(16))

Simplify Exponent: Simplify the exponent in the numerator. We have ((3^(2))^(-6)). According to the power of a power rule, we multiply the exponents. (3^(2 * -6)) = 3^(-12)Combine Exponents: Combine the exponents in the numerator and the denominator. We have 3^(-12) / 3^(4). According to the quotient of powers rule, we subtract the exponent in the denominator from the exponent in the numerator. 3^(-12 - 4) = 3^(-16)Write Final Expression: Write the final expression. The expression 3^(-16) can be written as 1 / 3^(16), which is the reciprocal of 3 raised to the 16th power.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Which expression is equivalent to
((3^(2))^(-6))/(3^(4))?

(1)/(3^(17))

3^(16)

(1)/(3^(18))

(1)/(3^(16))

Which expression is equivalent to $\frac{\left(3^{2}\right)^{-6}}{3^{4}} ?$ $\newline$ $\frac{1}{3^{17}}$ $\newline$ $3^{16}$ $\newline$ $\frac{1}{3^{18}}$ $\newline$ $\frac{1}{3^{16}}$

View step-by-step help

Home

Math Problems

Algebra 1

Multiplication with rational exponents

Full solution

Q. Which expression is equivalent to

\frac{\left(3^{2}\right)^{-6}}{3^{4}} ?

\newline

\frac{1}{3^{17}}

\newline

3^{16}

\newline

\frac{1}{3^{18}}

\newline

\frac{1}{3^{16}}

Simplify Exponent: Simplify the exponent in the numerator. $\newline$ We have $((3^{2})^{(-6)})$ . According to the power of a power rule, we multiply the exponents. $\newline$ $(3^{2 \times -6}) = 3^{-12}$
Combine Exponents: Combine the exponents in the numerator and the denominator. $\newline$ We have $3^{-12} / 3^{4}$ . According to the quotient of powers rule, we subtract the exponent in the denominator from the exponent in the numerator. $\newline$ $3^{-12 - 4} = 3^{-16}$
Write Final Expression: Write the final expression. $\newline$ The expression $3^{-16}$ can be written as $1 / 3^{16}$ , which is the reciprocal of $3$ raised to the $16$ th power.