How do you solve #(k ^ { 2} ) ^ { 2} \cdot 2k#?

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Answer 1 · 2017-04-17T15:06:24

See the entire solution process below:

First, use this rule of exponents to eliminate the term within parenthesis:

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

#(k^color(red)(2))^color(blue)(2) * 2k = k^(color(red)(2) xx color(blue)(2)) * 2k = k^4 * 2k#

We can then rewrite the expression as:

#2(k^4 * k)#

Now, use these two rules of exponents to complete the solution:

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

#2(k^4 * k) = 2(k^color(red)(4) * k^color(blue)(1)) = 2k^(color(red)(4) + color(blue)(1)) = 2k^5#