To solve the problem in the image, let's work through the steps. The task is to expand $5^{x+2}$ in ascending powers of $x$ up to the term in $x^3$. Here's the approach:

Step 1: Use the given property

The problem states that $5^x = e^{x \ln 5}$. So, we can write: $5^{x+2} = 5^x \cdot 5^2 = e^{x \ln 5} \cdot 25$

Step 2: Expand $e^{x \ln 5}$ using the Taylor series

The Taylor series for $e^y$ around $y = 0$ is: $e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \dots$ Here, $y = x \ln 5$. Substituting $y$, we get: $e^{x \ln 5} = 1 + (x \ln 5) + \frac{(x \ln 5)^2}{2!} + \frac{(x \ln 5)^3}{3!} + \dots$

Step 3: Multiply by $25$

Now, multiply the expanded series by $25$: $5^{x+2} = 25 \cdot \left(1 + (x \ln 5) + \frac{(x \ln 5)^2}{2} + \frac{(x \ln 5)^3}{6}\right)$ $5^{x+2} = 25 + 25 (x \ln 5) + \frac{25 (x \ln 5)^2}{2} + \frac{25 (x \ln 5)^3}{6}$

Step 4: Simplify each term

Distribute $25$ into each term: $5^{x+2} = 25 + 25 x \ln 5 + \frac{25 (\ln 5)^2 x^2}{2} + \frac{25 (\ln 5)^3 x^3}{6}$

Final Answer:

$5^{x+2} = 25 + 25 x \ln 5 + \frac{25 (\ln 5)^2 x^2}{2} + \frac{25 (\ln 5)^3 x^3}{6}$

Let me know if you'd like a deeper explanation of any part of the solution! Here are some related questions to further your understanding:

5 Related Questions:

1. How is the Taylor series for $e^y$ derived and used in expansions?
2. Why is the natural logarithm ($\ln$) used in this type of expansion for exponents?
3. How does truncating the series at a specific power of $x$ affect the accuracy of the approximation?
4. Can this method be generalized for other bases, like $7^{x+2}$?
5. What is the significance of the coefficient $25$ in the final expansion?

Tip:

Always verify that higher-order terms (beyond what’s required) are negligible for the desired level of precision in applications.