Explore a vast range of topics and get informed answers at IDNLearn.com. Receive prompt and accurate responses to your questions from our community of knowledgeable professionals ready to assist you at any time.
Sagot :
Let's solve the expression [tex]\(\log_{100 \sqrt{64}} \left(2^3 \times 512^{0.1}\right)\)[/tex] step-by-step.
1. Simplify the expression inside the logarithm:
[tex]\[ 2^3 \times 512^{0.1} \][/tex]
Firstly, notice that [tex]\(512\)[/tex] can be rewritten as a power of 2:
[tex]\[ 512 = 2^9 \][/tex]
Therefore,
[tex]\[ 512^{0.1} = (2^9)^{0.1} = 2^{9 \times 0.1} = 2^{0.9} \][/tex]
Now, substituting this back into the original expression:
[tex]\[ 2^3 \times 2^{0.9} = 2^{3+0.9} = 2^{3.9} \][/tex]
2. Simplify the base of the logarithm:
The base is given as [tex]\(100 \sqrt{64}\)[/tex].
We know that [tex]\(\sqrt{64} = 8\)[/tex],
So,
[tex]\[ 100 \sqrt{64} = 100 \times 8 = 800 \][/tex]
3. Rewrite the logarithm:
Now, our expression simplifies to:
[tex]\[ \log_{800} \left(2^{3.9}\right) \][/tex]
4. Using the change of base formula:
The change of base formula for logarithms is:
[tex]\[ \log_b(a) = \frac{\log(a)}{\log(b)} \][/tex]
Here, [tex]\(a = 2^{3.9}\)[/tex] and [tex]\(b = 800\)[/tex].
So,
[tex]\[ \log_{800} \left(2^{3.9}\right) = \frac{\log(2^{3.9})}{\log(800)} \][/tex]
5. Simplifying the numerator [tex]\(\log(2^{3.9})\)[/tex]:
We know from the properties of logarithms:
[tex]\[ \log(2^{3.9}) = 3.9 \cdot \log(2) \][/tex]
6. Calculate the value inside the logarithm:
The value of [tex]\(2^{3.9}\)[/tex] is approximately 14.929:
[tex]\[ 2^{3.9} \approx 14.93 \][/tex]
7. Compute the logarithms:
Given the result from calculations:
[tex]\[ \log(14.929) \approx 1.174 \][/tex]
[tex]\[ \log(800) \approx 2.905 \][/tex]
8. Combine the results together:
Finally:
[tex]\[ \frac{\log(14.929)}{\log(800)} = \frac{1.174}{2.905} \approx 0.404 \][/tex]
So, the value of [tex]\(\log_{800} \left(2^{3.9}\right)\)[/tex] is approximately 0.404.
Hence,
[tex]\[ \log_{100 \sqrt{64}}\left(2^3 \times 512^{0.1}\right) \approx 0.404 \][/tex]
Therefore, the value of the given logarithmic expression is approximately [tex]\(0.404\)[/tex].
1. Simplify the expression inside the logarithm:
[tex]\[ 2^3 \times 512^{0.1} \][/tex]
Firstly, notice that [tex]\(512\)[/tex] can be rewritten as a power of 2:
[tex]\[ 512 = 2^9 \][/tex]
Therefore,
[tex]\[ 512^{0.1} = (2^9)^{0.1} = 2^{9 \times 0.1} = 2^{0.9} \][/tex]
Now, substituting this back into the original expression:
[tex]\[ 2^3 \times 2^{0.9} = 2^{3+0.9} = 2^{3.9} \][/tex]
2. Simplify the base of the logarithm:
The base is given as [tex]\(100 \sqrt{64}\)[/tex].
We know that [tex]\(\sqrt{64} = 8\)[/tex],
So,
[tex]\[ 100 \sqrt{64} = 100 \times 8 = 800 \][/tex]
3. Rewrite the logarithm:
Now, our expression simplifies to:
[tex]\[ \log_{800} \left(2^{3.9}\right) \][/tex]
4. Using the change of base formula:
The change of base formula for logarithms is:
[tex]\[ \log_b(a) = \frac{\log(a)}{\log(b)} \][/tex]
Here, [tex]\(a = 2^{3.9}\)[/tex] and [tex]\(b = 800\)[/tex].
So,
[tex]\[ \log_{800} \left(2^{3.9}\right) = \frac{\log(2^{3.9})}{\log(800)} \][/tex]
5. Simplifying the numerator [tex]\(\log(2^{3.9})\)[/tex]:
We know from the properties of logarithms:
[tex]\[ \log(2^{3.9}) = 3.9 \cdot \log(2) \][/tex]
6. Calculate the value inside the logarithm:
The value of [tex]\(2^{3.9}\)[/tex] is approximately 14.929:
[tex]\[ 2^{3.9} \approx 14.93 \][/tex]
7. Compute the logarithms:
Given the result from calculations:
[tex]\[ \log(14.929) \approx 1.174 \][/tex]
[tex]\[ \log(800) \approx 2.905 \][/tex]
8. Combine the results together:
Finally:
[tex]\[ \frac{\log(14.929)}{\log(800)} = \frac{1.174}{2.905} \approx 0.404 \][/tex]
So, the value of [tex]\(\log_{800} \left(2^{3.9}\right)\)[/tex] is approximately 0.404.
Hence,
[tex]\[ \log_{100 \sqrt{64}}\left(2^3 \times 512^{0.1}\right) \approx 0.404 \][/tex]
Therefore, the value of the given logarithmic expression is approximately [tex]\(0.404\)[/tex].
We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. IDNLearn.com has the answers you need. Thank you for visiting, and we look forward to helping you again soon.