The expression you provided is \( ^4\log_{16}\sqrt{2} - ^4\log_{32} \).

Let's solve this step by step:

### Step 1: Rewrite the logarithms using the change of base formula
For any logarithm, we can use the change of base formula:
\[
^a\log_b c = \frac{\log_c c}{\log_c b}
\]

So, let's rewrite each logarithm:

\[
^4\log_{16}\sqrt{2} = \frac{\log_{4}\sqrt{2}}{\log_{4}16}
\]

\[
^4\log_{32} = \frac{\log_{4}32}{\log_{4}32}
\]

### Step 2: Simplify each logarithm
First, simplify \( ^4\log_{16}\sqrt{2} \):

1. \( 16 = 4^2 \) so \( \log_{4}16 = 2 \).
2. \( \sqrt{2} = 2^{1/2} \) so \( \log_{4}\sqrt{2} = \frac{1}{2}\log_{4}2 \).
3. Since \( 2 = 4^{1/2} \), then \( \log_{4}2 = \frac{1}{2} \).

Thus:

\[
^4\log_{16}\sqrt{2} = \frac{\frac{1}{2}}{2} = \frac{1}{4}
\]

Now, for \( ^4\log_{32} \):

1. \( 32 = 2^5 \) so \( \log_{4}32 = 5 \cdot \log_{4}2 = 5 \times \frac{1}{2} = \frac{5}{2} \).

Thus:

\[
^4\log_{32} = \frac{\frac{1}{2}}{\frac{5}{2}} = \frac{1}{5}
\]

### Step 3: Combine the results
Finally, subtract the results obtained:

\[
\frac{1}{4} - \frac{1}{5} = \frac{5 - 4}{20} = \frac{1}{20}
\]

### Final Answer
\[
\boxed{\frac{1}{20}}
\]

Would you like further details or have any questions?

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**Related Questions:**
1. How does the change of base formula work?
2. What are the properties of logarithms that simplify such problems?
3. How do you solve logarithmic equations with different bases?
4. How do you convert logarithmic expressions to different bases?
5. What is the connection between exponential and logarithmic functions?

**Tip:** When solving logarithmic expressions, always check if you can simplify using basic logarithm rules before applying complex operations.

Learn how to solve the logarithmic expression ^4log16sqrt(2) - ^4log32 step-by-step. This problem involves logarithms and the change of base formula, suitable for advanced high school students.

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