According to the property of the power rule for exponents, 

\(\displaystyle (x^a)^b = x^{a\times b}\)

The exponents may be multiplied if the exponent is outside of the parentheses.

\(\displaystyle (2a^{120})^{5} = 2^5 \cdot a^{120\times 5}\)

The answer is:  \(\displaystyle 32a^{600}\)

In order to simplify this, we will need to distribute the power of 20 across both powers inside the inner quantity.

\(\displaystyle (a^2b^{25})^{20} = (a^{2\cdot 20}b^{25\cdot 20})\)

Multiply the powers.

The answer is:  \(\displaystyle a^{40}b^{500}\)

Simplify the inner quantity first.

When powers of the same base are multiplied together, the powers may be added.

\(\displaystyle x^{35}\cdot x^{20} = x^{55}\)

The expression becomes:  \(\displaystyle (x^{55})^{15}\)

When the quantity of a power is raised to a certain power, the powers will need to be multiplied.

\(\displaystyle (x^{55})^{15} =x^{55\times 15} = x^{825}\)

The answer is:  \(\displaystyle x^{825}\)

The term can be simplified by distributing the power with the term inside the parentheses.

\(\displaystyle (2a^{300})^3 = 2^3\cdot a^{300\times 3} = 8a^{900}\)

The answer is:  \(\displaystyle 8a^{900}\)

Simplify the terms in the bracket first.  When a quantity of a power is raised to the power, the exponents may be multiplied.

\(\displaystyle [(a^{10})^5]^{20} =[(a^{10\times 5})]^{20} = [a^{50}]^{20}\)

Repeat the process for the term in the bracket.

\(\displaystyle a^{50\times 20}\)

The answer is:  \(\displaystyle a^{1000}\)

Simplify the terms inside the parentheses.

\(\displaystyle ( 3^3\cdot 3^2) = (3^{3+2}) = 3^5\)

\(\displaystyle ( 3^3\cdot 3^2)^{10} = (3^5)^{10}\)

The quantity will be multiplied by itself ten times.  This means that the powers here can be multiplied.

\(\displaystyle (3^5)^{10} = 3^{5\times 10} = 3^{50}\)

The answer is:  \(\displaystyle 3^{50}\)

Rewrite the second term so that it has the common base.

\(\displaystyle \frac{1}{36} = 6^{-2}\)

\(\displaystyle (6^{12})(\frac{1}{36})^{6} = (6^{12})(6^{-2})^{6}\)

Simplify the power outside of the parentheses by multiplication.

\(\displaystyle (6^{12})(6^{-12})\)

Since these exponents with the same bases are multiplied, the exponents can be added.

\(\displaystyle (6^{12})(6^{-12})= 6^{12+(-12)} = 6^0=1\)

The answer is:  \(\displaystyle 1\)

In order to simplify this, we will first need to simplify the inner term of the parentheses.

Rewrite the nine with base three.

\(\displaystyle 9=3^2\)

The expression becomes:

\(\displaystyle (3^2\cdot3^{12})^5\)

Since the common bases of a certain power are multiplied, the exponents can be added.

\(\displaystyle (3^2\cdot3^{12})^5=(3^{2+12})^5 = (3^{14})^5\)

Now that we have an exponent outside of a quantity, we can multiply the exponents together.

\(\displaystyle (3^{14})^5 = 3^{70}\)

The answer is:  \(\displaystyle 3^{70}\)

This expression can be simplified by either writing out the quantities four time, or simply by just multiplying the powers together.  Either method will give similar answers.

\(\displaystyle (6^9)^{4}=(6^9)(6^9)(6^9)(6^9)= 6^{9\times 4} = 6^{36}\)

Do not change the base at any point in the calculation.

The answer is:  \(\displaystyle 6^{36}\)

In order to solve this, use the distribution rule for exponents.

\(\displaystyle (a^{X})^Y = a^{XY}\)

Multiply the powers together.

\(\displaystyle (3^{25})^{25} = 3^{25\times 25} = 3^{625}\)

The answer is:  \(\displaystyle 3^{625}\)