Simplify 81$^{\frac{-3}{4}}$ x 25$^{\frac{1}{2}}$ x 243$^{\frac{2}{5}}$

To simplify the expression $81^{-\frac{3}{4}} \times 25^{\frac{1}{2}} \times 243^{\frac{2}{5}}$, we need to express each base as a power of its prime factors.

1. Analyze the components

• $81$: $81 = 3^4$
• $25$: $25 = 5^2$
• $243$: $243 = 3^5$

2. Substitute and apply the Power of a Power rule

The rule states that $(a^m)^n = a^{m \times n}$. Let's apply this to each term:

• First term: $(3^4)^{-\frac{3}{4}} = 3^{4 \times (-\frac{3}{4})} = 3^{-3}$ • Recall that $x^{-n} = \frac{1}{x^n}$, so $3^{-3} = \frac{1}{3^3} = \frac{1}{27}$

• Second term: $(5^2)^{\frac{1}{2}} = 5^{2 \times \frac{1}{2}} = 5^1 = 5$

• Third term: $(3^5)^{\frac{2}{5}} = 3^{5 \times \frac{2}{5}} = 3^2 = 9$

3. Combine the results

Now, multiply the simplified values together:

We can simplify this by multiplying $5 \times 9$ first:

Now, divide both 45 and 27 by their greatest common divisor, which is 9:

4. Conclusion

Comparing our result to the given options, we find that $\frac{5}{3}$ matches option D.

Exam Tip: When dealing with fractional exponents, always look for the prime base that allows the denominator of the exponent to cancel out. For example, seeing a denominator of "4" in the exponent should immediately make you look for a way to express the base as something to the power of 4 (like $81 = 3^4$). This makes the calculation much faster!