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The expression shown is:

\[
243^{\frac{7}{10}} \cdot 243^{\frac{1}{10}}
\]

To simplify this, we can use the **Product of Powers Rule**, which states:

\[
a^m \cdot a^n = a^{m+n}
\]

Here, both terms have the same base \( 243 \), so we can add the exponents:

\[
243^{\frac{7}{10} + \frac{1}{10}} = 243^{\frac{8}{10}} = 243^{\frac{4}{5}}
\]

Thus, the expression simplifies to \( 243^{\frac{4}{5}} \).

To compute \( 243^{\frac{4}{5}} \), we recall that the exponent \(\frac{4}{5}\) means the fifth root of 243 raised to the power of 4. First, we find the fifth root of 243:

\[
243^{\frac{1}{5}} = 3 \quad (\text{since } 3^5 = 243)
\]

Now raise 3 to the power of 4:

\[
3^4 = 81
\]

So, the final result is:

\[
243^{\frac{4}{5}} = 81
\]

The value of the expression is **81**.

Would you like a detailed explanation of any part? Here are 5 related questions you could explore:

1. How do you simplify expressions with fractional exponents?
2. What are the key exponent rules to know for similar problems?
3. How can you find roots of larger numbers efficiently?
4. What happens if the base is negative or a fraction in similar expressions?
5. How do exponents relate to logarithms in expressions like this?

**Tip:** When dealing with fractional exponents, remember that the denominator represents the root, and the numerator represents the power!

Learn how to simplify the expression 243^(7/10) * 243^(1/10) using the product of powers rule. This example demonstrates key algebraic concepts like fractional exponents and exponent rules, suitable for students in Grades 8-10.

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