Question

which of the following is equivalent to (3^x+1)(3^x-1)^2/9^x for all x? a. 3^2x-1 b. 3^x c. 3^x-1 d. 3^x-2 e. 3^3-x

Answer 1

`3^(x-1)`

Step-by-step explanation:

Given

`((3^(x+1))(3^(x-1))^2)/(9^x)`

Required

Find its equivalent

`((3^(x+1))(3^(x-1))^2)/(9^x)`

Expand the numerator

`((3^(x+1))(3^(x-1))(3^(x-1)))/(9^x)`

Apply law of indices:
`a^m * a^n = a^(m+n)`

This gives

`(3^(x+1+x-1+x-1))/(9^x)`

Collect like terms

`(3^(x+x+x+1-1-1))/(9^x)`

`(3^(3x-1))/(9^x)`

Express
`9^x` as a factor of 3

`(3^(3x-1))/(3^2^x)`

Apply law of indices:
`(a^m)/(a^n) = a^(m-n)`

This gives

`3^(3x-1-2x)`

`3^(3x-2x-1)`

`3^(x-1)`

Hence,

`((3^(x+1))(3^(x-1))^2)/(9^x)` is equivalent to
`3^(x-1)`