When expressing negative exponents, we rewrite as such: 

\(\displaystyle x^{-a}=\frac{1}{x^a}\) 

in which \(\displaystyle a\) is the positive exponent raising base \(\displaystyle x\).

\(\displaystyle (-5)^{-4}=\frac{1}{(-5)^4}=\frac{1}{625}\)

When expressing negative exponents, we rewrite as such: 

\(\displaystyle x^{-a}=\frac{1}{x^a}\) 

in which \(\displaystyle a\) is the positive exponent raising base \(\displaystyle x\).

\(\displaystyle \frac{1}{2}^{-3}=\frac{1}{(\frac{1}{2})^3}=\frac{1}{\frac{1}{8}}=8\)

When dealing with fractional exponents, we rewrite as such: 

\(\displaystyle x^\frac{a}{b}=\sqrt[b]{x^a}\) 

in which \(\displaystyle b\) is the index of the radical and \(\displaystyle a\) is the exponent raising base \(\displaystyle a\). 

When expressing negative exponents, we rewrite as such: 

\(\displaystyle x^{-a}=\frac{1}{x^a}\) 

in which \(\displaystyle a\) is the positive exponent raising base \(\displaystyle x\).

\(\displaystyle \frac{1}{6}^{-\frac{1}{2}}=\frac{1}{(\sqrt\frac{1}{6})}=\frac{1}{\frac{1}{\sqrt{6}}}=\sqrt{6}\)

When dealing with fractional exponents, we rewrite as such: 

\(\displaystyle x^\frac{a}{b}=\sqrt[b]{x^a}\) 

in which \(\displaystyle b\) is the index of the radical and \(\displaystyle a\) is the exponent raising base \(\displaystyle a\). 

When expressing negative exponents, we rewrite as such: 

\(\displaystyle x^{-a}=\frac{1}{x^a}\) 

in which \(\displaystyle a\) is the positive exponent raising base \(\displaystyle x\).

\(\displaystyle 12^{-\frac{1}{2}}=\frac{1}{(\sqrt{12})}=\frac{1}{2{\sqrt{3}}}=\frac{\sqrt{3}}{6}\) 

Remember when getting rid of radicals, just multiply top and bottom by that radical.

When dealing with fractional exponents, we rewrite as such: 

\(\displaystyle x^\frac{a}{b}=\sqrt[b]{x^a}\) 

in which \(\displaystyle b\) is the index of the radical and \(\displaystyle a\) is the exponent raising base \(\displaystyle a\). 

When expressing negative exponents, we rewrite as such: 

\(\displaystyle x^{-a}=\frac{1}{x^a}\) 

in which \(\displaystyle a\) is the positive exponent raising base \(\displaystyle x\).

\(\displaystyle 48^{-\frac{2}{3}}=\frac{1}{\sqrt[3]{48^2}}=\frac{1}{\sqrt[3]{2304}}\) 

Let's find a perfect cube which is 

\(\displaystyle 64 =(4^3)\).

\(\displaystyle \frac{1}{\sqrt[3]{2304}}=\frac{1}{\sqrt[3]{64}*\sqrt[3]{36}}=\frac{1}{4\sqrt[3]{36}}\) 

To simplify, we need to multiply top and bottom by an appropriate cubic root. We know \(\displaystyle 36=6^2\) so if we multiply top and bottom by \(\displaystyle \sqrt[3]{6}\) we will get an integer in the denominator.

\(\displaystyle \frac{1}{4\sqrt[3]{36}}*\frac{\sqrt[3]{6}}{\sqrt[3]{6}}=\frac{\sqrt[3]{6}}{4*6}=\frac{\sqrt[3]{6}}{24}\)

When dealing with fractional exponents, we rewrite as such: 

\(\displaystyle x^\frac{a}{b}=\sqrt[b]{x^a}\) 

in which \(\displaystyle b\) is the index of the radical and \(\displaystyle a\) is the exponent raising base \(\displaystyle a\). 

When expressing negative exponents, we rewrite as such: 

\(\displaystyle x^{-a}=\frac{1}{x^a}\) 

in which \(\displaystyle a\) is the positive exponent raising base \(\displaystyle x\).

\(\displaystyle 12^{-\frac{3}{4}}=\frac{1}{\sqrt[4]{12^3}}=\frac{1}{\sqrt[4]{1728}}\) 

Let's find a perfect fourth power which is 

\(\displaystyle 16 = (2^4)\).

\(\displaystyle \frac{1}{\sqrt[4]{1728}}=\frac{1}{\sqrt[4]{16}*\sqrt[4]{108}}=\frac{1}{2\sqrt[4]{108}}\) 

To simplify, we need to multiply top and bottom by an appropriate fourth root. We know \(\displaystyle 108=3^3*2^2\).

We need to complete the numbers to the fourth power. It we multiply top and bottom by \(\displaystyle \sqrt[4]{3*2^2}=\sqrt[4]{12}\) we will get an integer in the denominator.

\(\displaystyle \frac{1}{2\sqrt[4]{108}}*\frac{\sqrt[4]{12}}{\sqrt[4]{12}}=\frac{\sqrt[4]{12}}{2*6}=\frac{\sqrt[4]{12}}{12}\)

First multiply the like terms, remembering that when multiplying terms that have exponents, you add the exponents.

\(\displaystyle (4x^{2}y^{-6})(2x^{-7}y^{3})=(4\cdot 2)(x^{2}\cdot x^{-7})(y^{-6}\cdot y^{3})\)

\(\displaystyle =8x^{2-7}y^{-6+3}\)

\(\displaystyle =8x^{-5}y^{-3}\)

Negative exponents indicate that the term should be in the denominator, so the final answer is:

\(\displaystyle \frac{8}{x^{5}y^{3}}\)

Convert each negative exponent into fractional form.

\(\displaystyle a^{-n} = \frac{1}{a^n}\)

\(\displaystyle 6^{-2}+3^{-2}=\frac{1}{6^2}+\frac{1}{3^2}\)

Simplify the denominators.

\(\displaystyle \frac{1}{36}+\frac{1}{9}\)

Convert the second fraction with a common denominator of 36.

\(\displaystyle \frac{1}{36}+\frac{4}{36}= \frac{5}{36}\)

The answer is:  \(\displaystyle \frac{5}{36}\)

In order to determine the value of x, we will need to convert the base of the right side similar to the left.

Eight is similar to two cubed.  Rewrite the equation.

\(\displaystyle 2^{-3x}=2^3\)

Now that our bases are the same, we can set the exponents equal to each other.

\(\displaystyle -3x=3\)

Divide by negative three on both sides.

\(\displaystyle \frac{-3x}{-3}=\frac{3}{-3}\)

The answer is:  \(\displaystyle x=-1\)

The negative exponent can be converted into a fraction.

\(\displaystyle a^{-n} = \frac{1}{a^n}\)

Rewrite the fraction.

\(\displaystyle \frac{3x}{3^{-2}} = \frac{3x}{\frac{1}{3^{2}}}\)

Rewrite the complex fraction using a division sign.

\(\displaystyle \frac{3x}{\frac{1}{3^{2}}} = 3x\div \frac{1}{3^2}\)

Turn the division sign to a multiplication sign and take the reciprocal of the second term.

\(\displaystyle 3x\times \frac{3^2}{1}=3x(9) = 27x\)

The answer is:  \(\displaystyle 27x\)