How do you simplify #((3^6)^n times (81)^(2n)) / (3^n)^4#?

1 Answer
May 27, 2016

Answer:

# = color(green)(3^(10n)#

Explanation:

#((3^6)^n xx (81)^(2n)) /((3^n)^4#

  • Simplifying #81# by prime factorisation (expressing a number as a product of its prime factors):

#81 = 3 * 3 * 3 * 3 = color(blue)(3^4#

So, #81 ^(2n) = (3^4 )^(2n)#

  • Applying below mentioned property to the expression:
    #color(blue)(a^m)^n =a ^(mn)#
  • # (3^4 )^(2n) = color(green)( 3 ^(8n)#

  • #(3^6)^n = 3^(6n)#

  • #(3^n)^4 = 3^(4n)#

The expression can now be written as:

#((3^6)^n xx (81)^(2n)) /((3^n)^4 ) = (3^(6n) xx color(green)( 3 ^(8n))) /(3^ ( 4n)#

  • Applying below mentioned property to the numerator:
    #color(blue)(a^m xx a^n = a ^(m+n)#

# =3^((6n + 8n)) /(3^ ( 4n)#

# =3^(14n) /(3^ ( 4n)#

  • Applying below mentioned property to the expression:
    #color(blue)(a^m / a ^n = a ^(m-n)#

# =3^((14n - 4n))#

# = color(green)(3^(10n)#