Example 1: Find the value of $81^2 \times 3^{-7}$

Solution:

The goal is to express all terms with the same base (3) and use the Product Rule $a^m \times a^n = a^{m+n}$.

The value is 3.

**Example 2: Simplify $\frac{(25)^{\frac{3}{2}} \times (243)^{\frac{3}{5}}}{(16)^{\frac{5}{4}} \times (8)^{\frac{4}{3}}}$}

Solution:

Express all bases as powers of prime numbers ($25=5^2$, $243=3^5$, $16=2^4$, $8=2^3$).

The simplified value is $\frac{3375}{512}$.

**Example 3: Prove that $\left(\frac{64}{125}\right)^{-\frac{2}{3}} + \frac{1}{\left(\frac{256}{625}\right)^{\frac{1}{4}}} + \frac{\sqrt{25}}{\sqrt[3]{64}} = \frac{65}{16}$}

Solution (Left Hand Side – L.H.S.):

Break down each term and use the Negative Exponent of a Fraction Rule $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$ and the Fractional Exponent Rule $\sqrt[n]{a^m} = a^{\frac{m}{n}}$.

Term 1: $\left(\frac{64}{125}\right)^{-\frac{2}{3}}$

• $64 = 4^3$ and $125 = 5^3$.

• $\left(\frac{64}{125}\right)^{-\frac{2}{3}} = \left(\left(\frac{4}{5}\right)^3\right)^{-\frac{2}{3}} = \left(\frac{4}{5}\right)^{(3 \times -\frac{2}{3})} = \left(\frac{4}{5}\right)^{-2} = \left(\frac{5}{4}\right)^2$

Term 2: $\frac{1}{\left(\frac{256}{625}\right)^{\frac{1}{4}}}$

• $256 = 4^4$ and $625 = 5^4$.

• $\frac{1}{\left(\frac{256}{625}\right)^{\frac{1}{4}}} = \frac{1}{\left(\left(\frac{4}{5}\right)^4\right)^{\frac{1}{4}}} = \frac{1}{\left(\frac{4}{5}\right)^{(4 \times \frac{1}{4})}} = \frac{1}{\left(\frac{4}{5}\right)^1} = \frac{5}{4}$

Term 3: $\frac{\sqrt{25}}{\sqrt[3]{64}}$

• $\sqrt{25} = 5$

• $\sqrt[3]{64} = \sqrt[3]{4^3} = 4$

• $\frac{\sqrt{25}}{\sqrt[3]{64}} = \frac{5}{4}$

Combining the terms:

Since L.H.S. = $\frac{65}{16}$ and R.H.S. = $\frac{65}{16}$, the statement is proven.