Math worksheet for evaluating exponents with various problems involving fractions and negative exponents.

Problem: Evaluating Exponents

We are tasked with solving the given problems involving exponents. Let's solve each problem step by step.

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#### Problem 1:
$$
\left(\frac{5}{6}\right)^6 \times \left(\frac{5}{6}\right)^{-4}
$$

Solution:
When multiplying two expressions with the same base, we add their exponents:
$$
a^m \times a^n = a^{m+n}
$$
Here, the base is $\frac{5}{6}$, and the exponents are $6$ and $-4$. Adding the exponents:
$$
\left(\frac{5}{6}\right)^6 \times \left(\frac{5}{6}\right)^{-4} = \left(\frac{5}{6}\right)^{6 + (-4)} = \left(\frac{5}{6}\right)^2
$$
Thus, the answer is:
$$
\boxed{\left(\frac{5}{6}\right)^2}
$$

---

#### Problem 2:
$$
\left(\frac{7}{5}\right)^{-3} \times \left(\frac{7}{5}\right)^2
$$

Solution:
Again, when multiplying two expressions with the same base, we add their exponents:
$$
a^m \times a^n = a^{m+n}
$$
Here, the base is $\frac{7}{5}$, and the exponents are $-3$ and $2$. Adding the exponents:
$$
\left(\frac{7}{5}\right)^{-3} \times \left(\frac{7}{5}\right)^2 = \left(\frac{7}{5}\right)^{-3 + 2} = \left(\frac{7}{5}\right)^{-1}
$$
Using the property of negative exponents, $a^{-n} = \frac{1}{a^n}$:
$$
\left(\frac{7}{5}\right)^{-1} = \frac{1}{\left(\frac{7}{5}\right)} = \frac{5}{7}
$$
Thus, the answer is:
$$
\boxed{\frac{5}{7}}
$$

---

#### Problem 3:
$$
\left(-\frac{2}{3}\right)^{-3} \times \left(-\frac{2}{3}\right)^2
$$

Solution:
When multiplying two expressions with the same base, we add their exponents:
$$
a^m \times a^n = a^{m+n}
$$
Here, the base is $-\frac{2}{3}$, and the exponents are $-3$ and $2$. Adding the exponents:
$$
\left(-\frac{2}{3}\right)^{-3} \times \left(-\frac{2}{3}\right)^2 = \left(-\frac{2}{3}\right)^{-3 + 2} = \left(-\frac{2}{3}\right)^{-1}
$$
Using the property of negative exponents, $a^{-n} = \frac{1}{a^n}$:
$$
\left(-\frac{2}{3}\right)^{-1} = \frac{1}{-\frac{2}{3}} = -\frac{3}{2}
$$
Thus, the answer is:
$$
\boxed{-\frac{3}{2}}
$$

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#### Problem 4:
$$
\left\{\left(\frac{3}{2}\right)^{-2}\right\}^2
$$

Solution:
When raising a power to another power, we multiply the exponents:
$$
(a^m)^n = a^{m \cdot n}
$$
Here, the base is $\frac{3}{2}$, and the exponents are $-2$ and $2$. Multiplying the exponents:
$$
\left\{\left(\frac{3}{2}\right)^{-2}\right\}^2 = \left(\frac{3}{2}\right)^{-2 \cdot 2} = \left(\frac{3}{2}\right)^{-4}
$$
Using the property of negative exponents, $a^{-n} = \frac{1}{a^n}$:
$$
\left(\frac{3}{2}\right)^{-4} = \frac{1}{\left(\frac{3}{2}\right)^4} = \frac{1}{\frac{3^4}{2^4}} = \frac{2^4}{3^4} = \frac{16}{81}
$$
Thus, the answer is:
$$
\boxed{\frac{16}{81}}
$$

---

#### Problem 5:
$$
\left\{\left(\frac{1}{3}\right)^{-3} - \left(\frac{1}{3}\right)^{-3}\right\} \div \left(\frac{1}{4}\right)^{-3}
$$

Solution:
First, simplify the expression inside the curly braces:
$$
\left(\frac{1}{3}\right)^{-3} - \left(\frac{1}{3}\right)^{-3} = 0
$$
So the entire expression becomes:
$$
0 \div \left(\frac{1}{4}\right)^{-3}
$$
Any number divided by a non-zero number is zero. Therefore:
$$
0 \div \left(\frac{1}{4}\right)^{-3} = 0
$$
Thus, the answer is:
$$
\boxed{0}
$$

---

#### Problem 6:
$$
\left\{\left(\frac{4}{3}\right)^{-1} - \left(\frac{1}{4}\right)^{-1}\right\}^{-1}
$$

Solution:
First, simplify each term inside the curly braces:
- For $\left(\frac{4}{3}\right)^{-1}$, using the property of negative exponents:
$$
\left(\frac{4}{3}\right)^{-1} = \frac{1}{\frac{4}{3}} = \frac{3}{4}
$$
- For $\left(\frac{1}{4}\right)^{-1}$, using the property of negative exponents:
$$
\left(\frac{1}{4}\right)^{-1} = \frac{1}{\frac{1}{4}} = 4
$$
Now substitute these values back into the expression:
$$
\left\{\left(\frac{4}{3}\right)^{-1} - \left(\frac{1}{4}\right)^{-1}\right\} = \left\{\frac{3}{4} - 4\right\}
$$
To subtract these, express $4$ as a fraction with a denominator of $4$:
$$
4 = \frac{16}{4}
$$
So:
$$
\frac{3}{4} - 4 = \frac{3}{4} - \frac{16}{4} = \frac{3 - 16}{4} = \frac{-13}{4}
$$
Now, the expression becomes:
$$
\left\{\frac{-13}{4}\right\}^{-1}
$$
Using the property of negative exponents:
$$
\left(\frac{-13}{4}\right)^{-1} = \frac{1}{\frac{-13}{4}} = \frac{4}{-13} = -\frac{4}{13}
$$
Thus, the answer is:
$$
\boxed{-\frac{4}{13}}
$$

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#### Problem 7:
Find the value of $x$ in:
$$
\left(\frac{5}{3}\right)^{-4} \times \left(\frac{5}{3}\right)^{-5} = \left(\frac{5}{3}\right)^{3x}
$$

Solution:
When multiplying two expressions with the same base, we add their exponents:
$$
\left(\frac{5}{3}\right)^{-4} \times \left(\frac{5}{3}\right)^{-5} = \left(\frac{5}{3}\right)^{-4 + (-5)} = \left(\frac{5}{3}\right)^{-9}
$$
So the equation becomes:
$$
\left(\frac{5}{3}\right)^{-9} = \left(\frac{5}{3}\right)^{3x}
$$
Since the bases are the same, we can equate the exponents:
$$
-9 = 3x
$$
Solving for $x$:
$$
x = \frac{-9}{3} = -3
$$
Thus, the answer is:
$$
\boxed{-3}
$$

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#### Problem 8:
Find the value of $x$ in:
$$
\left(\frac{2}{9}\right)^4 \times \left(\frac{2}{9}\right)^7 = \left(\frac{2}{9}\right)^{2x}
$$

Solution:
When multiplying two expressions with the same base, we add their exponents:
$$
\left(\frac{2}{9}\right)^4 \times \left(\frac{2}{9}\right)^7 = \left(\frac{2}{9}\right)^{4 + 7} = \left(\frac{2}{9}\right)^{11}
$$
So the equation becomes:
$$
\left(\frac{2}{9}\right)^{11} = \left(\frac{2}{9}\right)^{2x}
$$
Since the bases are the same, we can equate the exponents:
$$
11 = 2x
$$
Solving for $x$:
$$
x = \frac{11}{2}
$$
Thus, the answer is:
$$
\boxed{\frac{11}{2}}
$$

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Final Answers:

1. $\boxed{\left(\frac{5}{6}\right)^2}$
2. $\boxed{\frac{5}{7}}$
3. $\boxed{-\frac{3}{2}}$
4. $\boxed{\frac{16}{81}}$
5. $\boxed{0}$
6. $\boxed{-\frac{4}{13}}$
7. $\boxed{-3}$
8. $\boxed{\frac{11}{2}}$