Answer:
[tex]3^{x-1}[/tex]
Explanation:
Given
[tex]\frac{(3^{x+1})(3^{x-1})^2}{9^x}[/tex]
Required
Find its equivalent
[tex]\frac{(3^{x+1})(3^{x-1})^2}{9^x}[/tex]
Expand the numerator
[tex]\frac{(3^{x+1})(3^{x-1})(3^{x-1})}{9^x}[/tex]
Apply law of indices: [tex]a^m * a^n = a^{m+n}[/tex]
This gives
[tex]\frac{3^{x+1+x-1+x-1}}{9^x}[/tex]
Collect like terms
[tex]\frac{3^{x+x+x+1-1-1}}{9^x}[/tex]
[tex]\frac{3^{3x-1}}{9^x}[/tex]
Express [tex]9^x[/tex] as a factor of 3
[tex]\frac{3^{3x-1}}{3^2^x}[/tex]
Apply law of indices: [tex]\frac{a^m}{a^n} = a^{m-n}[/tex]
This gives
[tex]3^{3x-1-2x}[/tex]
[tex]3^{3x-2x-1}[/tex]
[tex]3^{x-1}[/tex]
Hence,
[tex]\frac{(3^{x+1})(3^{x-1})^2}{9^x}[/tex] is equivalent to [tex]3^{x-1}[/tex]
