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To solve the equation [tex]\( 9^{n-2} + 2 \times 3^{2n-3} = 63 \)[/tex], we will follow a detailed step-by-step approach.
1. Express the equation in simpler terms:
- Notice that [tex]\( 9 \)[/tex] can be written as [tex]\( 3^2 \)[/tex].
- Therefore, [tex]\( 9^{n-2} \)[/tex] can be rewritten using the base 3 as [tex]\( (3^2)^{n-2} \)[/tex].
- Using the power rule [tex]\((a^m)^n = a^{mn}\)[/tex], we get [tex]\( 9^{n-2} = (3^2)^{n-2} = 3^{2(n-2)} \)[/tex].
2. Rewrite the original equation:
- Substitute [tex]\( 9^{n-2} \)[/tex] with [tex]\( 3^{2(n-2)} \)[/tex]:
[tex]\[ 3^{2(n-2)} + 2 \times 3^{2n-3} = 63 \][/tex]
3. Simplify the exponents:
- Simplify [tex]\( 3^{2(n-2)} \)[/tex]:
[tex]\[ 3^{2(n-2)} = 3^{2n-4} \][/tex]
- Thus, the equation becomes:
[tex]\[ 3^{2n-4} + 2 \times 3^{2n-3} = 63 \][/tex]
4. Factor out common terms:
- Notice that both terms have the base 3 raised to a power involving [tex]\(2n\)[/tex].
- Factor out the smallest exponent term, [tex]\( 3^{2n-4} \)[/tex]:
[tex]\[ 3^{2n-4} + 2 \times 3^{2n-3} = 3^{2n-4} \left(1 + 2 \times 3\right) \][/tex]
- Simplify inside the parentheses:
[tex]\[ 1 + 2 \times 3 = 1 + 6 = 7 \][/tex]
- So the equation is:
[tex]\[ 3^{2n-4} \times 7 = 63 \][/tex]
5. Isolate the term involving the exponent:
- Divide both sides of the equation by 7:
[tex]\[ 3^{2n-4} = \frac{63}{7} = 9 \][/tex]
6. Express 9 as a power of 3:
- Recall that [tex]\( 9 = 3^2 \)[/tex]:
[tex]\[ 3^{2n-4} = 3^2 \][/tex]
7. Match the exponents:
- Since the bases are identical, set the exponents equal to each other:
[tex]\[ 2n - 4 = 2 \][/tex]
8. Solve for [tex]\( n \)[/tex]:
- Add 4 to both sides:
[tex]\[ 2n = 6 \][/tex]
- Divide by 2:
[tex]\[ n = 3 \][/tex]
9. Verify the solution (if necessary):
- Substitute [tex]\( n = 3 \)[/tex] back into the original equation:
[tex]\[ 9^{3-2} + 2 \times 3^{2 \times 3 - 3} = 9^1 + 2 \times 3^3 = 9 + 2 \times 27 = 9 + 54 = 63 \][/tex]
- Since both sides of the equation are equal, [tex]\( n = 3 \)[/tex] is a valid solution.
In conclusion, the solution to the equation [tex]\( 9^{n-2} + 2 \times 3^{2n-3} = 63 \)[/tex] is [tex]\( n = 3 \)[/tex]. An alternate solution involves a complex number, represented as [tex]\(\frac{\log(27) + i\pi}{\log(3)}\)[/tex], but [tex]\( n = 3 \)[/tex] is the simplest and most straightforward answer.
1. Express the equation in simpler terms:
- Notice that [tex]\( 9 \)[/tex] can be written as [tex]\( 3^2 \)[/tex].
- Therefore, [tex]\( 9^{n-2} \)[/tex] can be rewritten using the base 3 as [tex]\( (3^2)^{n-2} \)[/tex].
- Using the power rule [tex]\((a^m)^n = a^{mn}\)[/tex], we get [tex]\( 9^{n-2} = (3^2)^{n-2} = 3^{2(n-2)} \)[/tex].
2. Rewrite the original equation:
- Substitute [tex]\( 9^{n-2} \)[/tex] with [tex]\( 3^{2(n-2)} \)[/tex]:
[tex]\[ 3^{2(n-2)} + 2 \times 3^{2n-3} = 63 \][/tex]
3. Simplify the exponents:
- Simplify [tex]\( 3^{2(n-2)} \)[/tex]:
[tex]\[ 3^{2(n-2)} = 3^{2n-4} \][/tex]
- Thus, the equation becomes:
[tex]\[ 3^{2n-4} + 2 \times 3^{2n-3} = 63 \][/tex]
4. Factor out common terms:
- Notice that both terms have the base 3 raised to a power involving [tex]\(2n\)[/tex].
- Factor out the smallest exponent term, [tex]\( 3^{2n-4} \)[/tex]:
[tex]\[ 3^{2n-4} + 2 \times 3^{2n-3} = 3^{2n-4} \left(1 + 2 \times 3\right) \][/tex]
- Simplify inside the parentheses:
[tex]\[ 1 + 2 \times 3 = 1 + 6 = 7 \][/tex]
- So the equation is:
[tex]\[ 3^{2n-4} \times 7 = 63 \][/tex]
5. Isolate the term involving the exponent:
- Divide both sides of the equation by 7:
[tex]\[ 3^{2n-4} = \frac{63}{7} = 9 \][/tex]
6. Express 9 as a power of 3:
- Recall that [tex]\( 9 = 3^2 \)[/tex]:
[tex]\[ 3^{2n-4} = 3^2 \][/tex]
7. Match the exponents:
- Since the bases are identical, set the exponents equal to each other:
[tex]\[ 2n - 4 = 2 \][/tex]
8. Solve for [tex]\( n \)[/tex]:
- Add 4 to both sides:
[tex]\[ 2n = 6 \][/tex]
- Divide by 2:
[tex]\[ n = 3 \][/tex]
9. Verify the solution (if necessary):
- Substitute [tex]\( n = 3 \)[/tex] back into the original equation:
[tex]\[ 9^{3-2} + 2 \times 3^{2 \times 3 - 3} = 9^1 + 2 \times 3^3 = 9 + 2 \times 27 = 9 + 54 = 63 \][/tex]
- Since both sides of the equation are equal, [tex]\( n = 3 \)[/tex] is a valid solution.
In conclusion, the solution to the equation [tex]\( 9^{n-2} + 2 \times 3^{2n-3} = 63 \)[/tex] is [tex]\( n = 3 \)[/tex]. An alternate solution involves a complex number, represented as [tex]\(\frac{\log(27) + i\pi}{\log(3)}\)[/tex], but [tex]\( n = 3 \)[/tex] is the simplest and most straightforward answer.
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