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Question
Simplify and hence find the value of n:
`(3^(2"n")*9^2*3^(-"n"))/(3^(3"n"))` = 27
Solution
Given
`(3^(2"n")*9^2*3^(-"n"))/(3^(3"n"))` = 27
`(3^(2"n" - "n")*(3^2)^2)/(3^(3"n"))` = 33
`3^"n"*(3^2)^2*3^(-3"n")` = 33
`3^"n"*3^4*3^(-3"n")` = 33
`3^("n" + 4 - 3"n")` = 33
`3^(4 - 2"n")` = 33
4 – 2n = 3
2n = 4 – 3
2n = 1
n = `1/2`
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