Math Problem Statement

⁴Log16√2 - ⁴Log32

Solution

The expression you provided is 4log1624log32^4\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: alogbc=logcclogcb^a\log_b c = \frac{\log_c c}{\log_c b}

So, let's rewrite each logarithm:

4log162=log42log416^4\log_{16}\sqrt{2} = \frac{\log_{4}\sqrt{2}}{\log_{4}16}

4log32=log432log432^4\log_{32} = \frac{\log_{4}32}{\log_{4}32}

Step 2: Simplify each logarithm

First, simplify 4log162^4\log_{16}\sqrt{2}:

  1. 16=4216 = 4^2 so log416=2\log_{4}16 = 2.
  2. 2=21/2\sqrt{2} = 2^{1/2} so log42=12log42\log_{4}\sqrt{2} = \frac{1}{2}\log_{4}2.
  3. Since 2=41/22 = 4^{1/2}, then log42=12\log_{4}2 = \frac{1}{2}.

Thus:

4log162=122=14^4\log_{16}\sqrt{2} = \frac{\frac{1}{2}}{2} = \frac{1}{4}

Now, for 4log32^4\log_{32}:

  1. 32=2532 = 2^5 so log432=5log42=5×12=52\log_{4}32 = 5 \cdot \log_{4}2 = 5 \times \frac{1}{2} = \frac{5}{2}.

Thus:

4log32=1252=15^4\log_{32} = \frac{\frac{1}{2}}{\frac{5}{2}} = \frac{1}{5}

Step 3: Combine the results

Finally, subtract the results obtained:

1415=5420=120\frac{1}{4} - \frac{1}{5} = \frac{5 - 4}{20} = \frac{1}{20}

Final Answer

120\boxed{\frac{1}{20}}

Would you like further details or have any questions?


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.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Change of Base Formula

Formulas

Change of Base Formula: ^a\log_b c = \frac{\log_c c}{\log_c b}

Theorems

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Suitable Grade Level

Advanced High School