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Exponents Rules

1. The 4 Important Exponents Rules

Here is a brief introduction to the four important rules for exponents in algebra:

Product of Powers Rule: When multiplying two terms with the same base, we add the exponents.
Example with numbers: \(2^3 \cdot 2^4 = 2^{3+4} = 2^7\)
Example with variables: \(x^3 \cdot x^4 = x^{3+4} = x^7\)

Quotient of Powers Rule: When dividing two terms with the same base, we subtract the exponents.
Example with numbers: \(\dfrac{3^6}{3^2} = 3^{6-2} = 3^4\)
Example with variables: \(\dfrac{y^6}{y^2} = y^{6-2} = y^4\)

Power of a Power Rule: When you have a power raised to another power, we multiply the exponents.
Example with numbers: \((2^3)^4 = 2^{3 \cdot 4} = 2^{12}\)
Example with variables: \((z^3)^4 = z^{3 \cdot 4} = z^{12}\)
Zero Power Rule: Any nonzero number raised to the power of zero is 1.
Example with numbers: \(7^0 = 1\)
Example with variables: \(a^0 = 1\)

Remember, understanding and applying these rules will help you simplify and solve more complex algebraic expressions and equations. Be careful to ensure the bases are the same when applying these rules.

2. Negative Exponents

Negative exponents can be understood by using the idea of reciprocals and the properties of exponents.

When a number or variable is raised to a negative exponent, it indicates that the reciprocal of that number or variable should be taken to the positive exponent.

Let's take an example: \(\frac{1}{{4^{-3}}}\)

To simplify this expression, we can rewrite it as the reciprocal with a positive exponent:

\[\frac{1}{{4^{-3}}} = \frac{1}{{\frac{1}{{4^3}}}}\]

Now, let's simplify the expression within the parentheses:

\(\frac{1}{{4^3}}\)

Here, \(4^3\) means \(4 \times 4 \times 4\), which is equal to 64.

Therefore, we have:

\[\frac{1}{{\frac{1}{{4^3}}}} = \frac{1}{{\frac{1}{64}}} = 64\]

So, \(\frac{1}{{4^{-3}}} = 64\).

In this case, the negative exponent flips the fraction and changes the sign of the exponent. It allows us to convert a negative exponent into a positive exponent by taking the reciprocal of the base.

3. Simplifying Exponent Expressions

Solve This:
Evaluate this expression \(\dfrac{5y^3}{\left(y^{-2}\right)^3}\)

To solve this problem, follow the order of operations and apply the exponent rules.

First, simplify the expression within the parentheses. To raise a power to another power, we multiply the exponents:
\[\left(y^{-2}\right)^3 = y^{-2 \times 3} = y^{-6}\]

Now, rewrite the original expression. To divide by a negative exponent, rewrite it as a positive exponent:
\[\frac{5y^3}{y^{-6}} = \frac{5y^3}{\frac{1}{y^6}}\]

Dividing by a fraction is equivalent to multiplying by its reciprocal:
\[5y^3 \times \frac{y^6}{1} = 5y^{3 + 6} = 5y^9\]

Therefore, the simplified expression is \(5y^9\).

4. Solving Equations with Exponents


Solve the equation: \[ 7^{4a+3} = 7^{a+9}\]

To solve the equation, we can equate the exponents:
\[4a+3 = a+9 \Rightarrow 3a + 3 = 9\]

Solving for \(a\):
\[3a = 6 \Rightarrow a = 2\]

Therefore, the solution to the equation is \(a = 2\).

Solve This:
If \(16^{20} = 2^x\), then find the value of \(x\).

We know that \(16 = 2^4\) because \(2 \times 2 \times 2 \times 2 = 16\).

Using the property of exponents, we can rewrite the equation as \((2^4)^{20} = 2^x\).

Applying the power rule, we multiply the exponents:

\[2^{4 \times 20} = 2^x\]

Simplifying the exponent on the left side, we have:

\[2^{80} = 2^x \Rightarrow 80 = x\]

Therefore, in the equation \(16^{20} = 2^x\), the value of \(x\) is 80.


Solve This:
If \(4a - b = 2\), find the value of \(\frac{{81^a}}{{3^b}}\)

We are given that \(4a - b = 2\).

Let's simplify the expression \(\frac{{81^a}}{{3^b}}\) using the properties of exponents.

We know that \(81 = 3^4\) because \(3 \times 3 \times 3 \times 3 = 81\).

Substituting this into the expression, and using the power of a power rule, we multiply the exponents we have:

\[\frac{{(3^4)^a}}{{3^b}} = 3^{4a - b}\]

Since we are given that \(4a - b = 2\), we can substitute it back into the expression:

\[3^{4a - b} = 3^2\]

Therefore, the value of \(\frac{{81^a}}{{3^b}}\) when \(4a - b = 2\) is 9.

5. Convert Radicals to Fractional Exponents


Radicals can be represented using fractional exponents, providing a more concise and flexible notation. Here are a few examples:

  1. Square Root: The square root of a number can be expressed as the number raised to the power of \(1/2\). For instance, \(\sqrt{5}\) is equivalent to \(5^{1/2}\).

  2. Cube Root: The cube root of a number can be represented as the number raised to the power of \(1/3\). For example, \(\sqrt[3]{5}\) is equal to \(5^{1/3}\).

  3. Higher Roots: The same concept applies to higher roots. To find the seventh root of \(10^3\), we write it as \(\sqrt[7]{10^3}\), which simplifies to \(10^{3/7}\).

\[ \sqrt{5} = 5^{1/2} \] \[ \sqrt[3]{5} = 5^{1/3} \] \[ \sqrt[7]{10^3} = 10^{3/7} \]

Converting radicals to fractional exponents allows for easier calculations and manipulation of expressions involving roots.

Solve This:
If \(\sqrt[7]{10^5} \cdot \sqrt[7]{10^9} = 10^x\), find \(x\).

To solve this problem, we'll simplify each radical expression:

\[\sqrt[7]{10^5} = (10^5)^{\frac{1}{7}} = 10^{\frac{5}{7}}\]
\[\sqrt[7]{10^9} = (10^9)^{\frac{1}{7}} = 10^{\frac{9}{7}}\]

Now, we can multiply the simplified radicals:

\[\sqrt[7]{10^5} \cdot \sqrt[7]{10^9} = 10^{\frac{5}{7}} \cdot 10^{\frac{9}{7}}\]

To multiply exponential expressions with the same base, we add the exponents:

\[10^{\frac{5}{7}} \cdot 10^{\frac{9}{7}} = 10^{\frac{5}{7} + \frac{9}{7}}\]

Combining the fractions:

\[10^{\frac{5+9}{7}} = 10^{\frac{14}{7}}\]

Simplifying the exponent:

\[10^{\frac{14}{7}} = 10^2\]

Therefore, the expression \(\sqrt[7]{10^5} \cdot \sqrt[7]{10^9}\) simplifies to \(10^2\).

Since this expression is equal to \(10^x\), we can equate the exponents:

\[x = 2\]

Hence, the value of \(x\) in the equation \(10^x\) is \(2\).

Solve This:
If \(2\sqrt{x+3} = \sqrt{8x}\), find \(x\).

To solve this equation, we will isolate the square root terms and then square both sides:

\[2\sqrt{x+3} = \sqrt{8x}\]
\[(2\sqrt{x+3})^2 = (\sqrt{8x})^2\]
\[4(x+3) = 8x\]

Expanding and simplifying the equation:

\[4x + 12 = 8x\]

Bringing like terms together:

\[4x - 8x = -12\]

Combining like terms:

\[-4x = -12\]

Dividing both sides by \(-4\) to solve for \(x\):

\[x = 3\]

Hence, the solution to the equation \(2\sqrt{x+3} = \sqrt{8x}\) is \(x = 3\).