How do we write #(sqrt7)^3# in exponential form? Algebra Exponents and Exponential Functions Exponential Properties Involving Products 1 Answer Shwetank Mauria Jul 20, 2016 #(sqrt7)^3=7^(3/2)# Explanation: To write #(sqrt7)^3# in exponential form, one needs two identities - #root(n)a=a^(1/n)# and #(a^m)^n=a^((m×n))# which is true for all rational #m# and #n#. Hence, #(sqrt7)^3# = #(7^(1/2))^3# = #7^((1/2×3))# = #7^(3/2)# Answer link Related questions What is the Exponential Property Involving Products? How do you apply the "product of powers" property to simplify expressions? What is an exponent and exponential notation? What is the difference between #-5^2# and #(-5)^2# ? How do you write 3(-2a)(-2a)(-2a)(-2a)# in exponential notation? How do you simplify #2^2 \cdot 2^4 \cdot 2^6#? How do you simplify #(4a^2)(-3a)(-5a^4)# using the product of powers property? How do you simplify #(-2xy^4z^2)^5#? How do you apply the exponential properties to simplify #(-8x)^3(5x)^2#? How do you write the prime factorization of 280? See all questions in Exponential Properties Involving Products Impact of this question 4640 views around the world You can reuse this answer Creative Commons License