Discover new perspectives and gain insights with IDNLearn.com. Our Q&A platform offers detailed and trustworthy answers to ensure you have the information you need.
Sagot :
To solve the given equation
[tex]\[ \frac{9^n \cdot 3^2 \cdot 3^n - 27^n}{(3^m \cdot 2)^{27}} = 3^{-3} \][/tex]
let's start by simplifying both the numerator and the denominator.
First, simplify the expressions in the numerator:
1. [tex]\(9^n\)[/tex] can be written as [tex]\((3^2)^n = 3^{2n}\)[/tex].
2. [tex]\(3^2\)[/tex] is [tex]\(9\)[/tex] or [tex]\(3^2\)[/tex].
3. [tex]\(3^n\)[/tex] is itself [tex]\(3^n\)[/tex].
4. [tex]\(27^n\)[/tex] can be written as [tex]\((3^3)^n = 3^{3n}\)[/tex].
Using these equivalences, the numerator becomes:
[tex]\[ 9^n \cdot 3^2 \cdot 3^n - 27^n = 3^{2n} \cdot 3^2 \cdot 3^n - 3^{3n} \][/tex]
Next, combine the exponents of 3 in the simplified numerator:
[tex]\[ 3^{2n} \cdot 3^2 \cdot 3^n = 3^{(2n + 2 + n)} = 3^{(3n + 2)} \][/tex]
Thus, the numerator simplifies to:
[tex]\[ 3^{3n + 2} - 3^{3n} \][/tex]
Now, simplify the denominator:
[tex]\[ (3^m \cdot 2)^{27} \][/tex]
Raise each term to the power of 27:
[tex]\[ 3^{27m} \cdot 2^{27} \][/tex]
Now the equation can be rewritten as:
[tex]\[ \frac{3^{3n + 2} - 3^{3n}}{3^{27m} \cdot 2^{27}} = 3^{-3} \][/tex]
For the right-hand side to equal [tex]\(3^{-3}\)[/tex], the terms involving powers of 3 in the numerator and denominator must be consistent in their exponents.
To equate the exponents of 3, let's express the powers clearly.
From the left-hand side, focus on the most significant terms involving powers of 3. Collect the exponent terms:
[tex]\[ 3^{3n + 2} - 3^{3n} \][/tex]
If we factor out [tex]\(3^{3n}\)[/tex] from the numerator:
[tex]\[ 3^{3n}(3^2 - 1) = 3^{3n} \cdot 8 = 3^{3n} \cdot 2^3 \][/tex]
Thus, the equation becomes:
[tex]\[ \frac{3^{3n} \cdot 2^3}{3^{27m} \cdot 2^{27}} = 3^{-3} \][/tex]
Now, simplifying the fractions:
[tex]\[ \frac{3^{3n} \cdot 2^3}{3^{27m} \cdot 2^{27}} = \frac{3^{3n}}{3^{27m}} \cdot \frac{2^3}{2^{27}} = 3^{3n-27m} \cdot 2^{3-27} \][/tex]
We know this equals [tex]\(3^{-3}\)[/tex]. So that gives us two separate equations for the exponents. The exponent of 3:
[tex]\[ 3n - 27m = -3 \][/tex]
The exponent of 2:
[tex]\[ 3 - 27 = -24 \][/tex]
These two should hold true. For the exponent of 3:
[tex]\[ 3n - 27m = -3 \implies n - 9m = -1 \implies m - n = 1 \][/tex]
Therefore, we have shown that:
[tex]\[ m - n = 1 \][/tex]
[tex]\[ \frac{9^n \cdot 3^2 \cdot 3^n - 27^n}{(3^m \cdot 2)^{27}} = 3^{-3} \][/tex]
let's start by simplifying both the numerator and the denominator.
First, simplify the expressions in the numerator:
1. [tex]\(9^n\)[/tex] can be written as [tex]\((3^2)^n = 3^{2n}\)[/tex].
2. [tex]\(3^2\)[/tex] is [tex]\(9\)[/tex] or [tex]\(3^2\)[/tex].
3. [tex]\(3^n\)[/tex] is itself [tex]\(3^n\)[/tex].
4. [tex]\(27^n\)[/tex] can be written as [tex]\((3^3)^n = 3^{3n}\)[/tex].
Using these equivalences, the numerator becomes:
[tex]\[ 9^n \cdot 3^2 \cdot 3^n - 27^n = 3^{2n} \cdot 3^2 \cdot 3^n - 3^{3n} \][/tex]
Next, combine the exponents of 3 in the simplified numerator:
[tex]\[ 3^{2n} \cdot 3^2 \cdot 3^n = 3^{(2n + 2 + n)} = 3^{(3n + 2)} \][/tex]
Thus, the numerator simplifies to:
[tex]\[ 3^{3n + 2} - 3^{3n} \][/tex]
Now, simplify the denominator:
[tex]\[ (3^m \cdot 2)^{27} \][/tex]
Raise each term to the power of 27:
[tex]\[ 3^{27m} \cdot 2^{27} \][/tex]
Now the equation can be rewritten as:
[tex]\[ \frac{3^{3n + 2} - 3^{3n}}{3^{27m} \cdot 2^{27}} = 3^{-3} \][/tex]
For the right-hand side to equal [tex]\(3^{-3}\)[/tex], the terms involving powers of 3 in the numerator and denominator must be consistent in their exponents.
To equate the exponents of 3, let's express the powers clearly.
From the left-hand side, focus on the most significant terms involving powers of 3. Collect the exponent terms:
[tex]\[ 3^{3n + 2} - 3^{3n} \][/tex]
If we factor out [tex]\(3^{3n}\)[/tex] from the numerator:
[tex]\[ 3^{3n}(3^2 - 1) = 3^{3n} \cdot 8 = 3^{3n} \cdot 2^3 \][/tex]
Thus, the equation becomes:
[tex]\[ \frac{3^{3n} \cdot 2^3}{3^{27m} \cdot 2^{27}} = 3^{-3} \][/tex]
Now, simplifying the fractions:
[tex]\[ \frac{3^{3n} \cdot 2^3}{3^{27m} \cdot 2^{27}} = \frac{3^{3n}}{3^{27m}} \cdot \frac{2^3}{2^{27}} = 3^{3n-27m} \cdot 2^{3-27} \][/tex]
We know this equals [tex]\(3^{-3}\)[/tex]. So that gives us two separate equations for the exponents. The exponent of 3:
[tex]\[ 3n - 27m = -3 \][/tex]
The exponent of 2:
[tex]\[ 3 - 27 = -24 \][/tex]
These two should hold true. For the exponent of 3:
[tex]\[ 3n - 27m = -3 \implies n - 9m = -1 \implies m - n = 1 \][/tex]
Therefore, we have shown that:
[tex]\[ m - n = 1 \][/tex]
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Your questions deserve precise answers. Thank you for visiting IDNLearn.com, and see you again soon for more helpful information.
