Answer:
The option "StartFraction 1 Over 3 Superscript 8" is correct
That is [tex]\frac{1}{3^8}[/tex] is correct answer
Therefore [tex][(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}[/tex]
Step-by-step explanation:
Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared
The given expression can be written as
[tex][(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2[/tex]
To find the simplified form of the given expression :
[tex][(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2[/tex]
[tex]=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2[/tex] ( using the property [tex](ab)^m=a^m.b^m[/tex] )
[tex]=(2^6)(3^{-12})\times (2^{-6})(3^4)[/tex] ( using the property [tex](a^m)^n=a^{mn}[/tex]
[tex]=(2^6)(2^{-6})(3^{-12})(3^4)[/tex] ( combining the like powers )
[tex]=2^{6-6}3^{-12+4}[/tex] ( using the property [tex]a^m.a^n=a^{m+n}[/tex] )
[tex]=2^03^{-8}[/tex]
[tex]=\frac{1}{3^8}[/tex] ( using the property [tex]a^{-m}=\frac{1}{a^m}[/tex] )
Therefore [tex][(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}[/tex]
Therefore option "StartFraction 1 Over 3 Superscript 8" is correct
That is [tex]\frac{1}{3^8}[/tex] is correct answer
