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The problem is asking to calculate Emre's followers on 15.06.2024, starting from 15.06.2020, where the number of followers increases by 1/3 each year.

### Information provided:
- Initial number of followers on 15.06.2020 = \( 15^4 \)
- Follower growth rate: 1/3 increase each year.

This is a case of exponential growth where the number of followers increases by a factor of \( 1 + \frac{1}{3} = \frac{4}{3} \) each year.

The formula for this type of growth is:

\[
N(t) = N_0 \times \left(\frac{4}{3}\right)^t
\]

Where:
- \( N_0 = 15^4 \) (initial number of followers),
- \( t = 4 \) years (from 2020 to 2024),
- \( \frac{4}{3} \) is the yearly growth factor.

Let's calculate the number of followers on 15.06.2024.The number of followers Emre will have on 15.06.2024 is approximately 160,000.

Would you like a more detailed explanation or have any other questions?

Here are 5 related questions to further your understanding:
1. How would the result change if the growth rate was 50% per year instead of 33%?
2. What would the number of followers be after 10 years, given the same growth rate?
3. Can you derive a general formula for the number of followers after any number of years?
4. What if the growth rate changes each year? How would that affect the calculation?
5. How does compound growth differ from linear growth in such scenarios?

**Tip**: Exponential growth occurs in many real-world situations, from finance to biology, where rates compound over time.

Emre is an active social media user. From the date of 15.06.2020, Emre has become a social media influencer by increasing his follower count by 1/3 each year compared to the previous year. If the number of followers on 15.06.2020 was 15^4, how many followers were there on 15.06.2024?

Learn how to calculate the exponential growth of a social media follower count. Starting with 15^4 followers in 2020 and increasing by 1/3 each year, the number of followers by 2024 is approximately 160,000. This problem covers key exponential growth concepts and algebraic formulas. Suitable for Grades 10-12 students.

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