How do you simplify #(4xy)^-1# and write it using only positive exponents?

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Answer 1 · 2017-03-25T20:08:02

#1/(4xy)#

Permit me to explain using examples

#color(blue)("Example 1")#

Suppose we had #b^2/b#

This is the same as #(bxxb)/b " "=" "b/bxxb" "=" "1xxb=b#

Going back to the beginning: Write as #b^2/b^1#

No consider #b^(2-1)=b^1=b# so it works to subtract the powers

#color(blue)("Example 2")#

Suppose we had #b/b^2#

This is the same as #b/(bxxb)" "=" "b/bxx1/b = color(red)(1/b)#

No using the method from above

#b/b^2" "=" "b^(1-2) " "=" "color(red)(b^(-1))#

So #b^(-1) = 1/b" "# and using the same principle #b^(-2)=1/b^2#

#color(blue)("Example 3")#

Not going through the whole demonstration process just accept that: #1/b^(-2)=b^2#
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#color(blue)("Answering the question")#

Using the principles demonstrated above. The power of -1 is applied to everything inside the brackets.

So #(4xy)^(-1) = 1/(4xy)#