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See attached photo

 

 Apr 17, 2020
 #1
avatar+23555 
+1

A little difficult to type in all of those exponents, so here is a hand-scrawled version:

 Apr 17, 2020
 #2
avatar+9447 
+1

See attached photo

 

Hello Guest!

 

\(\color{BrickRed}[\frac{(2a^{-3}b^4)^2}{(3a^5b)^{-2}}]^{-4}\ badly\ readable\\ =[\frac{4a^{-6}b^8}{3^{-2}a^{-10}b^{-2}}]^{-4}\\ =\frac{3^{-8}a^{-40}b^{-8}}{4^4a^{-40}b^{-8}}\)

\(=\frac{1}{3^8\cdot4^4}\)

\(=\frac{1}{1679616}\)

laugh  !

\(\color{BrickRed}[\frac{(2a^{-3}b^4)^2}{(3a^5b)^{-2}}]^{-1}\)\(=\frac{1}{36}\)

 Apr 17, 2020
edited by asinus  Apr 17, 2020
 #3
avatar+109501 
+1

\(\left( \frac{(2a^{-3}b^4)^2}{(3a^5b)^{-2}} \right)^{-1}\\ = \frac{(3a^5b)^{-2}} {(2a^{-3}b^4)^2} \\ = \frac{1} {(2a^{-3}b^4)^2 (3a^5b)^{2}} \\ = \frac{1} {(4a^{-6}b^8) (9a^{10}b^2)} \\ = \frac{1} {36a^{4}b^{10}} \\\)

 

Which is the same as EP's answer  :)

 Apr 17, 2020

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