https://file.aiquickdraw.com/mathbot/file/1728918388592kx7gu9ea.jpg

The equation in the image is:

\[
49^{1 + \log_7{4}} - \log_7{\frac{1}{49}}
\]

Let's break it down step by step.

### 1. Simplifying the first part: \( 49^{1 + \log_7{4}} \)
- We know that \( 49 = 7^2 \). So, rewrite it as:
  \[
  49^{1 + \log_7{4}} = (7^2)^{1 + \log_7{4}} = 7^{2(1 + \log_7{4})}
  \]
- Now distribute the 2:
  \[
  7^{2(1 + \log_7{4})} = 7^{2 + 2\log_7{4}}
  \]
- Using the logarithmic identity \( a\log_b{c} = \log_b{c^a} \), we can simplify:
  \[
  7^{2 + \log_7{16}} = 7^{2 + \log_7{16}} = 7^{\log_7{(16 \cdot 7^2)}} = 7^{\log_7{784}}
  \]
- So, the first term simplifies to:
  \[
  7^{\log_7{784}} = 784
  \]

### 2. Simplifying the second part: \( \log_7{\frac{1}{49}} \)
- We can rewrite \( \frac{1}{49} \) as \( 49^{-1} = (7^2)^{-1} = 7^{-2} \).
- Using the logarithmic rule \( \log_b{a^c} = c\log_b{a} \), we get:
  \[
  \log_7{7^{-2}} = -2
  \]

### 3. Putting it all together:
- The expression becomes:
  \[
  784 - (-2) = 784 + 2 = 786
  \]

The final simplified result is:

\[
\boxed{786}
\]

Would you like further details or have any questions? Here are some related questions to deepen understanding:

1. How would the result change if the logarithm base were different?
2. Can you apply these simplifications to more complex logarithmic functions?
3. What is the general process for simplifying expressions with logarithms and exponents?
4. How does the property \( a\log_b{c} = \log_b{c^a} \) work in other scenarios?
5. Could this type of problem be solved graphically or by using a calculator?

**Tip**: Always try to express everything in terms of the same base when dealing with exponent and logarithm combinations!

Simplify the expression: 49^{1 + \log_7{4}} - \log_7{\frac{1}{49}}.

Learn how to simplify the expression 49^{1 + log7(4)} - log7(1/49) by applying logarithmic identities and exponentiation rules. This detailed solution covers logarithmic properties, exponentiation, and simplification techniques for advanced algebra students. Ideal for Grade 11-12 students.

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