How do you multiply #6v ^ { 8} \cdot 3y ^ { 4} \cdot 4x ^ { 6} y ^ { - 4} v ^ { - 9} x#?

1 Answer
May 20, 2017

See a solution process below:

Explanation:

First, rewrite this expression as:

#(6 * 3 * 4)(x^6 * x)(y^4 * y^-4)(v^8 * v^-9) =>#

#72(x^6 * x)(y^4 * y^-4)(v^8 * v^-9)#

Next, use these rules of exponents to multiply the #x# terms:

#a = a^color(blue)(1)# and #x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a) + color(blue)(b))#

#72(x^6 * x)(y^4 * y^-4)(v^8 * v^-9) =>#

#72(x^color(red)(6) * x^color(blue)(1))(y^4 * y^-4)(v^8 * v^-9) =>#

#72x^(color(red)(6) + color(blue)(1))(y^4 * y^-4)(v^8 * v^-9) =>#

#72x^7(y^4 * y^-4)(v^8 * v^-9)#

Then, use these rules of exponents to multiply the #y# terms:

#x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a) + color(blue)(b))# and #a^color(red)(0) = 1#

#72x^7(y^color(red)(4) * y^color(blue)(-4))(v^8 * v^-9) =>#

#72x^7y^(color(red)(4) + color(blue)(-4))(v^8 * v^-9) =>#

#72x^7y^color(red)(0)(v^8 * v^-9) =>#

#72x^7 1(v^8 * v^-9) =>#

#72x^7(v^8 * v^-9)#

Now, use these rules of exponents to multiply the #v# terms:

#x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a) + color(blue)(b))# and #x^color(red)(a) = 1/x^color(red)(-a)# and #a^color(red)(1) = a#

#72x^7(v^color(red)(8) * v^color(blue)(-9)) =>#

#72x^7v^(color(red)(8) + color(blue)(-9)) =>#

#72x^7v^color(red)(-1) =>#

#(72x^7)/v^color(red)(- -1) =>#

#(72x^7)/v^color(red)(1) =>#

#(72x^7)/v#