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Calculate the following expression:
[tex]\log _{100 \sqrt{64}}\left(2^3 \times 512^{0 \cdot 1}\right)[/tex]
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Response:


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].
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