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Sagot :
Certainly! Let's solve the problem step-by-step in a detailed manner:
1. Identify the Components:
- We have a number: [tex]\( 9 \)[/tex]
- This number is raised to a power: [tex]\( 6 \)[/tex]
- The base of the logarithm we are considering: [tex]\( 3 \)[/tex]
2. Logarithm of a Number Raised to a Power:
- We start with the expression: [tex]\( \log_3(9^6) \)[/tex]
- The property of logarithms that applies here is: [tex]\( \log_b(a^c) = c \cdot \log_b(a) \)[/tex]
- According to this property, we can rewrite [tex]\( \log_3(9^6) \)[/tex] as:
[tex]\[ \log_3(9^6) = 6 \cdot \log_3(9) \][/tex]
3. Simplify the Expression [tex]\( \log_3(9) \)[/tex]:
- Next, we need to evaluate [tex]\( \log_3(9) \)[/tex]
- Recall that [tex]\( 9 \)[/tex] is [tex]\( 3^2 \)[/tex]. Therefore, [tex]\( \log_3(9) \)[/tex] is the same as [tex]\( \log_3(3^2) \)[/tex]
- Using the property of logarithms again, [tex]\( \log_b(a^c) = c \cdot \log_b(a) \)[/tex], we have:
[tex]\[ \log_3(3^2) = 2 \cdot \log_3(3) \][/tex]
4. Evaluate [tex]\( \log_3(3) \)[/tex]:
- The logarithm of a number to its own base is [tex]\( 1 \)[/tex]. Thus, [tex]\( \log_3(3) = 1 \)[/tex]
- So, substituting this value, we get:
[tex]\[ \log_3(3^2) = 2 \cdot 1 = 2 \][/tex]
5. Combine the Results:
- Substituting [tex]\( \log_3(9) = 2 \)[/tex] back into our original expression:
[tex]\[ \log_3(9^6) = 6 \cdot \log_3(9) = 6 \cdot 2 = 12 \][/tex]
6. Conclusion:
- Therefore, the solution to the problem [tex]\( \log_3(9^6) \)[/tex] is:
[tex]\[ \log_3(9^6) = 12 \][/tex]
So, we have verified that the logarithm of a number raised to a power is the same as the power times the logarithm of the number, and in this problem, it accurately translates to the result:
[tex]\[ \boxed{12} \][/tex]
1. Identify the Components:
- We have a number: [tex]\( 9 \)[/tex]
- This number is raised to a power: [tex]\( 6 \)[/tex]
- The base of the logarithm we are considering: [tex]\( 3 \)[/tex]
2. Logarithm of a Number Raised to a Power:
- We start with the expression: [tex]\( \log_3(9^6) \)[/tex]
- The property of logarithms that applies here is: [tex]\( \log_b(a^c) = c \cdot \log_b(a) \)[/tex]
- According to this property, we can rewrite [tex]\( \log_3(9^6) \)[/tex] as:
[tex]\[ \log_3(9^6) = 6 \cdot \log_3(9) \][/tex]
3. Simplify the Expression [tex]\( \log_3(9) \)[/tex]:
- Next, we need to evaluate [tex]\( \log_3(9) \)[/tex]
- Recall that [tex]\( 9 \)[/tex] is [tex]\( 3^2 \)[/tex]. Therefore, [tex]\( \log_3(9) \)[/tex] is the same as [tex]\( \log_3(3^2) \)[/tex]
- Using the property of logarithms again, [tex]\( \log_b(a^c) = c \cdot \log_b(a) \)[/tex], we have:
[tex]\[ \log_3(3^2) = 2 \cdot \log_3(3) \][/tex]
4. Evaluate [tex]\( \log_3(3) \)[/tex]:
- The logarithm of a number to its own base is [tex]\( 1 \)[/tex]. Thus, [tex]\( \log_3(3) = 1 \)[/tex]
- So, substituting this value, we get:
[tex]\[ \log_3(3^2) = 2 \cdot 1 = 2 \][/tex]
5. Combine the Results:
- Substituting [tex]\( \log_3(9) = 2 \)[/tex] back into our original expression:
[tex]\[ \log_3(9^6) = 6 \cdot \log_3(9) = 6 \cdot 2 = 12 \][/tex]
6. Conclusion:
- Therefore, the solution to the problem [tex]\( \log_3(9^6) \)[/tex] is:
[tex]\[ \log_3(9^6) = 12 \][/tex]
So, we have verified that the logarithm of a number raised to a power is the same as the power times the logarithm of the number, and in this problem, it accurately translates to the result:
[tex]\[ \boxed{12} \][/tex]
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