We are tasked with solving the equation \( \log_4(32 + 4^x) \).

Let's break it down step by step:

1. **Rewrite the logarithmic equation**:
   The logarithmic equation \( \log_b(a) = c \) means \( a = b^c \). In this case:
   \[
   \log_4(32 + 4^x) = y
   \]
   implies:
   \[
   32 + 4^x = 4^y
   \]
   
2. **Express \(4^x\) and \(4^y\)**:
   Both \(4^x\) and \(4^y\) can be expressed as powers of 2. Recall that \( 4 = 2^2 \), so:
   \[
   4^x = (2^2)^x = 2^{2x}
   \]
   and
   \[
   4^y = (2^2)^y = 2^{2y}
   \]
   Thus, the equation becomes:
   \[
   32 + 2^{2x} = 2^{2y}
   \]
   
3. **Simplify 32**:
   Since \(32 = 2^5\), we now have:
   \[
   2^5 + 2^{2x} = 2^{2y}
   \]
   
   At this point, we have to solve this equation for \(x\).

Let me know if you would like to proceed further with solving this equation step by step!

Explore the solution to the logarithmic equation log4(32 + 4^x) using properties of logarithms and exponents. This detailed step-by-step breakdown covers essential algebraic concepts like expressing logarithmic terms and simplifying exponential expressions. Ideal for students in Grades 9-12.

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