How do you simplify #(sqrt49(a^2)(b^-8))#?

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Answer 1 · 2015-10-12T23:07:09

#(7a^2)/b^8#

Your starting expression looks like this

#sqrt(49) * a^2 * b^(-8)#

The first thing to notice here is that #49# is a perfect square, which means that you can write

#49 = 7 * 7 = 7^2#

#sqrt(49) = sqrt(7^2) = 7#

The expression becomes

#7 * a^2 * b^(-8)#

Next, rewrite the negative exponent by using

#x^(-n) = 1/x^n" "#, for #x!=0#

In your case, you have

#b^(-8) = 1/b^8#

The expression will thus be

#sqrt(49) * a^2 * b^(-8) = color(green)(7 * a^2 * 1/b^8)#