Question
Evaluate log(10^11⋅10^5) .(1 point)
Answers
Answer
Use the rule logb(xy)=log bx+log by to find log4(16⋅64) .(1 point)
Answer
If log3 41≈3.38 and log32≈0.631 , what is log3 82 , to the nearest thousandth?(1 point)
Answer
If log12≈1.0792 and log3≈0.4771 , find log36 to four decimal places.(1 point) log36≈
Answered by
GPT-4o mini
To evaluate \( \log(10^{11} \cdot 10^{5}) \), we can use the property of logarithms that states:
\[
\log(a \cdot b) = \log(a) + \log(b)
\]
In this case, we can rewrite the logarithm as follows:
\[
\log(10^{11} \cdot 10^{5}) = \log(10^{11}) + \log(10^{5})
\]
Now, we can apply another property of logarithms:
\[
\log(10^x) = x
\]
Thus, we have:
\[
\log(10^{11}) = 11 \quad \text{and} \quad \log(10^{5}) = 5
\]
Now we can substitute these back into our equation:
\[
\log(10^{11} \cdot 10^{5}) = 11 + 5 = 16
\]
Therefore, the result is:
\[
\boxed{16}
\]
\[
\log(a \cdot b) = \log(a) + \log(b)
\]
In this case, we can rewrite the logarithm as follows:
\[
\log(10^{11} \cdot 10^{5}) = \log(10^{11}) + \log(10^{5})
\]
Now, we can apply another property of logarithms:
\[
\log(10^x) = x
\]
Thus, we have:
\[
\log(10^{11}) = 11 \quad \text{and} \quad \log(10^{5}) = 5
\]
Now we can substitute these back into our equation:
\[
\log(10^{11} \cdot 10^{5}) = 11 + 5 = 16
\]
Therefore, the result is:
\[
\boxed{16}
\]
Answered by
GPT-4o mini
To find \( \log_4(16 \cdot 64) \) using the property \( \log_b(xy) = \log_b(x) + \log_b(y) \), we can first rewrite it as follows:
\[
\log_4(16 \cdot 64) = \log_4(16) + \log_4(64)
\]
Next, we will calculate \( \log_4(16) \) and \( \log_4(64) \).
1. **Calculating \( \log_4(16) \)**:
- Since \( 16 = 4^2 \), we have:
\[
\log_4(16) = \log_4(4^2) = 2
\]
2. **Calculating \( \log_4(64) \)**:
- Since \( 64 = 4^3 \) (because \( 4^3 = (2^2)^3 = 2^6 = 64 \)), we have:
\[
\log_4(64) = \log_4(4^3) = 3
\]
Now we can substitute these values back into our logarithmic expression:
\[
\log_4(16 \cdot 64) = \log_4(16) + \log_4(64) = 2 + 3 = 5
\]
Therefore, the result is:
\[
\boxed{5}
\]
\[
\log_4(16 \cdot 64) = \log_4(16) + \log_4(64)
\]
Next, we will calculate \( \log_4(16) \) and \( \log_4(64) \).
1. **Calculating \( \log_4(16) \)**:
- Since \( 16 = 4^2 \), we have:
\[
\log_4(16) = \log_4(4^2) = 2
\]
2. **Calculating \( \log_4(64) \)**:
- Since \( 64 = 4^3 \) (because \( 4^3 = (2^2)^3 = 2^6 = 64 \)), we have:
\[
\log_4(64) = \log_4(4^3) = 3
\]
Now we can substitute these values back into our logarithmic expression:
\[
\log_4(16 \cdot 64) = \log_4(16) + \log_4(64) = 2 + 3 = 5
\]
Therefore, the result is:
\[
\boxed{5}
\]
Answered by
GPT-4o mini
To find \( \log_3 82 \) using the given logarithms \( \log_3 41 \approx 3.38 \) and \( \log_3 2 \approx 0.631 \), we can express \( 82 \) in terms of \( 41 \) and \( 2 \):
\[
82 = 41 \cdot 2
\]
Using the property of logarithms \( \log_b(xy) = \log_b(x) + \log_b(y) \), we have:
\[
\log_3 82 = \log_3 (41 \cdot 2) = \log_3 41 + \log_3 2
\]
Substituting the known values:
\[
\log_3 82 \approx 3.38 + 0.631
\]
Now, we add these values together:
\[
\log_3 82 \approx 3.38 + 0.631 \approx 4.011
\]
Rounding to the nearest thousandth, we get:
\[
\log_3 82 \approx \boxed{4.011}
\]
\[
82 = 41 \cdot 2
\]
Using the property of logarithms \( \log_b(xy) = \log_b(x) + \log_b(y) \), we have:
\[
\log_3 82 = \log_3 (41 \cdot 2) = \log_3 41 + \log_3 2
\]
Substituting the known values:
\[
\log_3 82 \approx 3.38 + 0.631
\]
Now, we add these values together:
\[
\log_3 82 \approx 3.38 + 0.631 \approx 4.011
\]
Rounding to the nearest thousandth, we get:
\[
\log_3 82 \approx \boxed{4.011}
\]
Answered by
GPT-4o mini
To find \( \log 36 \), we can first express \( 36 \) in terms of its prime factors:
\[
36 = 6^2 = (2 \cdot 3)^2 = 2^2 \cdot 3^2
\]
Thus, we have:
\[
\log 36 = \log(2^2 \cdot 3^2) = \log(2^2) + \log(3^2) = 2\log 2 + 2\log 3
\]
We need to find \( \log 2 \). Since we know \( \log 12 \) and can express \( 12 \) as \( 4 \cdot 3 \), or \( 2^2 \cdot 3 \), we can use this to find \( \log 2 \):
\[
\log 12 = \log(2^2 \cdot 3) = 2\log 2 + \log 3
\]
We know that \( \log 12 \approx 1.0792 \) and \( \log 3 \approx 0.4771 \). Substituting these values in:
\[
1.0792 = 2\log 2 + 0.4771
\]
Now, we can solve for \( \log 2 \):
\[
2\log 2 = 1.0792 - 0.4771
\]
\[
2\log 2 = 0.6021
\]
\[
\log 2 = \frac{0.6021}{2} = 0.30105
\]
Now that we have \( \log 2 \), we can find \( \log 36 \):
\[
\log 36 = 2\log 2 + 2\log 3 = 2(0.30105) + 2(0.4771)
\]
\[
= 0.6021 + 0.9542 = 1.5563
\]
Therefore, rounding to four decimal places, we have:
\[
\log 36 \approx \boxed{1.5563}
\]
\[
36 = 6^2 = (2 \cdot 3)^2 = 2^2 \cdot 3^2
\]
Thus, we have:
\[
\log 36 = \log(2^2 \cdot 3^2) = \log(2^2) + \log(3^2) = 2\log 2 + 2\log 3
\]
We need to find \( \log 2 \). Since we know \( \log 12 \) and can express \( 12 \) as \( 4 \cdot 3 \), or \( 2^2 \cdot 3 \), we can use this to find \( \log 2 \):
\[
\log 12 = \log(2^2 \cdot 3) = 2\log 2 + \log 3
\]
We know that \( \log 12 \approx 1.0792 \) and \( \log 3 \approx 0.4771 \). Substituting these values in:
\[
1.0792 = 2\log 2 + 0.4771
\]
Now, we can solve for \( \log 2 \):
\[
2\log 2 = 1.0792 - 0.4771
\]
\[
2\log 2 = 0.6021
\]
\[
\log 2 = \frac{0.6021}{2} = 0.30105
\]
Now that we have \( \log 2 \), we can find \( \log 36 \):
\[
\log 36 = 2\log 2 + 2\log 3 = 2(0.30105) + 2(0.4771)
\]
\[
= 0.6021 + 0.9542 = 1.5563
\]
Therefore, rounding to four decimal places, we have:
\[
\log 36 \approx \boxed{1.5563}
\]