Answer:
The values of a and b are a=3 and b=3.
Step-by-step explanation:
Given fractional equation is [tex]\frac{(2xy)^4}{4xy}=4x^ay^b[/tex]
To find the values of a and b to make given equation true :
[tex]\frac{(2xy)^4}{4xy}=4x^ay^b[/tex]
Take LHS [tex]\frac{(2xy)^4}{4xy}[/tex]
[tex]=\frac{2^4x^4y^4}{4xy}[/tex] ( using the property [tex](ab)^m=a^mb^m[/tex] )
[tex]=\frac{16x^4y^4}{4xy}[/tex]
[tex]=4x^4y^4x^{-1}y^{-1}[/tex] ( using the property [tex]\frac{1}{a^m}=a^{-m}[/tex] )
[tex]=4x^{4-1}y^{4-1}[/tex] ( using the property [tex]a^m.a^n=a^{m+n}[/tex] )
[tex]=4x^3y^3[/tex]
[tex]\frac{(2xy)^4}{4xy}=4x^3y^3[/tex]
Comparing with [tex]\frac{(2xy)^4}{4xy}=4x^ay^b=4x^3y^3[/tex]
Therefore [tex]4x^ay^b=4x^3y^3[/tex]
Therefore the values of a and b are a=3 and b=3
