https://file.aiquickdraw.com/mathbot/file/1728925350454bg3dmm19.jpg

The equation provided in the image is:

\[
21 \cdot 16^6 + 11 \cdot 8^8 = 2^{3n-1}
\]

To solve for \( n \), let's break down the process step by step:

### Step 1: Simplify the left-hand side
We'll first calculate the values of \( 16^6 \) and \( 8^8 \) to make it easier.

- \( 16^6 = (2^4)^6 = 2^{24} \)
- \( 8^8 = (2^3)^8 = 2^{24} \)

Now, substitute these into the equation:

\[
21 \cdot 2^{24} + 11 \cdot 2^{24} = 2^{3n-1}
\]

### Step 2: Factor the left-hand side
Since both terms have \( 2^{24} \) in them, we can factor it out:

\[
(21 + 11) \cdot 2^{24} = 2^{3n-1}
\]

Simplifying further:

\[
32 \cdot 2^{24} = 2^{3n-1}
\]

### Step 3: Express \( 32 \) as a power of 2
We know that \( 32 = 2^5 \), so we can rewrite the equation as:

\[
2^5 \cdot 2^{24} = 2^{3n-1}
\]

This simplifies to:

\[
2^{29} = 2^{3n-1}
\]

### Step 4: Equate the exponents
Since the bases are the same, we can equate the exponents:

\[
29 = 3n - 1
\]

### Step 5: Solve for \( n \)
Now, solve for \( n \):

\[
29 + 1 = 3n \implies 30 = 3n \implies n = \frac{30}{3} = 10
\]

Thus, the value of \( n \) is \( \boxed{10} \).

---

Would you like more details on any of these steps? 

Here are 5 related questions to further explore:

1. How would you approach solving an equation with different powers of two on both sides?
2. What is the process for factoring powers of two in equations?
3. Can you explain how exponents are combined when multiplying numbers with the same base?
4. What is the significance of rewriting terms in powers of two in these types of problems?
5. How does understanding the properties of exponents help in simplifying complex equations?

**Tip**: Always look for common factors in terms involving exponents, as it can simplify the process significantly.

21·16^6 + 11·8^8 = 2^(3n-1) olduğuna göre, n kaçtır?

Solve the equation 21·16^6 + 11·8^8 = 2^(3n-1) step-by-step by simplifying powers of 2. Learn how to apply exponentiation rules and factor powers of 2 to find the value of n. Suitable for high school algebra students.

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