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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

Simplify and hence find the value of n: nnn32n⋅92⋅3-n33n = 27 - Mathematics

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प्रश्न

Simplify and hence find the value of n:

`(3^(2"n")*9^2*3^(-"n"))/(3^(3"n"))` = 27

बेरीज

उत्तर

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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पाठ 2: Basic Algebra - Exercise 2.11 [पृष्ठ ७७]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 2 Basic Algebra
Exercise 2.11 | Q 4 | पृष्ठ ७७
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