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To solve the equation [tex]\(\frac{9^n \times 3^2 \times\left(3^{\frac{-n}{2}}\right)^{-2}-(27)^n}{3^m \times 2^3}=\frac{1}{27}\)[/tex] for [tex]\(m - n\)[/tex], let's proceed with the following steps:
1. Simplify the given expression [tex]\(\frac{9^n \times 3^2 \times \left(3^{\frac{-n}{2}}\right)^{-2} - (27)^n}{3^m \times 2^3}\)[/tex].
Let's break it down.
2. Note that [tex]\(9^n\)[/tex] can be written as [tex]\((3^2)^n\)[/tex], which simplifies to [tex]\(3^{2n}\)[/tex].
3. The term [tex]\( (3^{\frac{-n}{2}})^{-2} \)[/tex] can be simplified using the properties of exponents: [tex]\( (a^b)^c = a^{bc} \)[/tex]. So [tex]\((3^{\frac{-n}{2}})^{-2}\)[/tex] becomes [tex]\(3^{\left(\frac{-n}{2} \times -2\right)} = 3^n\)[/tex].
4. The term [tex]\(3^2\)[/tex] remains as [tex]\(3^2\)[/tex].
5. The term [tex]\((27)^n\)[/tex] can be written as [tex]\((3^3)^n\)[/tex], which simplifies to [tex]\(3^{3n}\)[/tex].
Now substitute these into the original equation:
[tex]\[ \frac{(3^{2n}) \times 3^2 \times 3^n - 3^{3n}}{3^m \times 2^3} = \frac{1}{27} \][/tex]
6. Combine like terms in the numerator:
[tex]\( 3^{2n} \times 3^2 \times 3^n = 3^{n+2n+2} = 3^{3n+2} \)[/tex]
So the expression becomes:
[tex]\[ \frac{3^{3n+2} - 3^{3n}}{3^m \times 2^3} = \frac{1}{27} \][/tex]
7. Simplify the numerator:
Factor out [tex]\(3^{3n}\)[/tex]:
[tex]\[ 3^{3n+2} - 3^{3n} = 3^{3n}(3^2 - 1) = 3^{3n} \times 8 \][/tex]
This provides:
[tex]\[ \frac{8 \times 3^{3n}}{3^m \times 2^3} = \frac{1}{27} \][/tex]
8. Note that [tex]\(2^3 = 8\)[/tex]:
[tex]\[ \frac{8 \times 3^{3n}}{3^m \times 8} = \frac{1}{27} \][/tex]
Simplify the 8:
[tex]\[ \frac{3^{3n}}{3^m} = \frac{1}{27} \][/tex]
9. Use properties of exponents: [tex]\( \frac{a^m}{a^n} = a^{m-n} \)[/tex]:
[tex]\[ 3^{3n - m} = \frac{1}{27} \][/tex]
10. Recall that [tex]\(27 = 3^3\)[/tex] and [tex]\(\frac{1}{27} = 3^{-3}\)[/tex]:
[tex]\[ 3^{3n - m} = 3^{-3} \][/tex]
11. Equate the exponents since the bases are the same:
[tex]\[ 3n - m = -3 \][/tex]
Solve for [tex]\(m\)[/tex]:
[tex]\[ m = 3n + 3 \][/tex]
12. Therefore, [tex]\(m - n = 3n + 3 - n = 2n + 3\)[/tex].
Thus, the answer is:
[tex]\[ \boxed{2n + 3} \][/tex]
So, if we match this with the given options:
[tex]\(m - n = 2n + 3 \)[/tex].
The correct answer is not directly one of the given choices (B, C, or D), but our detailed algebraic solution shows the result. None of the provided options exactly matches [tex]\(2n + 3\)[/tex], so we may need to reconsider the context or check if there has been a transcription or typographical error in the original problem statement or provided answers.
1. Simplify the given expression [tex]\(\frac{9^n \times 3^2 \times \left(3^{\frac{-n}{2}}\right)^{-2} - (27)^n}{3^m \times 2^3}\)[/tex].
Let's break it down.
2. Note that [tex]\(9^n\)[/tex] can be written as [tex]\((3^2)^n\)[/tex], which simplifies to [tex]\(3^{2n}\)[/tex].
3. The term [tex]\( (3^{\frac{-n}{2}})^{-2} \)[/tex] can be simplified using the properties of exponents: [tex]\( (a^b)^c = a^{bc} \)[/tex]. So [tex]\((3^{\frac{-n}{2}})^{-2}\)[/tex] becomes [tex]\(3^{\left(\frac{-n}{2} \times -2\right)} = 3^n\)[/tex].
4. The term [tex]\(3^2\)[/tex] remains as [tex]\(3^2\)[/tex].
5. The term [tex]\((27)^n\)[/tex] can be written as [tex]\((3^3)^n\)[/tex], which simplifies to [tex]\(3^{3n}\)[/tex].
Now substitute these into the original equation:
[tex]\[ \frac{(3^{2n}) \times 3^2 \times 3^n - 3^{3n}}{3^m \times 2^3} = \frac{1}{27} \][/tex]
6. Combine like terms in the numerator:
[tex]\( 3^{2n} \times 3^2 \times 3^n = 3^{n+2n+2} = 3^{3n+2} \)[/tex]
So the expression becomes:
[tex]\[ \frac{3^{3n+2} - 3^{3n}}{3^m \times 2^3} = \frac{1}{27} \][/tex]
7. Simplify the numerator:
Factor out [tex]\(3^{3n}\)[/tex]:
[tex]\[ 3^{3n+2} - 3^{3n} = 3^{3n}(3^2 - 1) = 3^{3n} \times 8 \][/tex]
This provides:
[tex]\[ \frac{8 \times 3^{3n}}{3^m \times 2^3} = \frac{1}{27} \][/tex]
8. Note that [tex]\(2^3 = 8\)[/tex]:
[tex]\[ \frac{8 \times 3^{3n}}{3^m \times 8} = \frac{1}{27} \][/tex]
Simplify the 8:
[tex]\[ \frac{3^{3n}}{3^m} = \frac{1}{27} \][/tex]
9. Use properties of exponents: [tex]\( \frac{a^m}{a^n} = a^{m-n} \)[/tex]:
[tex]\[ 3^{3n - m} = \frac{1}{27} \][/tex]
10. Recall that [tex]\(27 = 3^3\)[/tex] and [tex]\(\frac{1}{27} = 3^{-3}\)[/tex]:
[tex]\[ 3^{3n - m} = 3^{-3} \][/tex]
11. Equate the exponents since the bases are the same:
[tex]\[ 3n - m = -3 \][/tex]
Solve for [tex]\(m\)[/tex]:
[tex]\[ m = 3n + 3 \][/tex]
12. Therefore, [tex]\(m - n = 3n + 3 - n = 2n + 3\)[/tex].
Thus, the answer is:
[tex]\[ \boxed{2n + 3} \][/tex]
So, if we match this with the given options:
[tex]\(m - n = 2n + 3 \)[/tex].
The correct answer is not directly one of the given choices (B, C, or D), but our detailed algebraic solution shows the result. None of the provided options exactly matches [tex]\(2n + 3\)[/tex], so we may need to reconsider the context or check if there has been a transcription or typographical error in the original problem statement or provided answers.
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