Problem 54 - Algebra and Trigonometry

Question

\(47-72\) . Simplify the expression, and eliminate any negative exponent(s). $$ \left(2 u^{2} v^{3}\right)^{3}\left(3 u^{-3} v\right)^{2} $$

Step-by-Step Solution

Answer

The simplified expression is \(72v^{11}\).

Step 1: Expand the First Expression

Expand the first expression by distributing the cube: \[ (2u^2v^3)^3 = 2^3(u^2)^3(v^3)^3 = 8u^6v^9 \]

Step 2: Expand the Second Expression

Expand the second expression by distributing the square: \[ (3u^{-3}v)^2 = 3^2(u^{-3})^2(v)^2 = 9u^{-6}v^2 \]

Step 3: Multiply Expanded Expressions

Multiply the expanded expressions together: \[ 8u^6v^9 \times 9u^{-6}v^2 \] Combine like terms: \[ = 8 \times 9 \times u^{6 + (-6)} \times v^{9 + 2} \] This simplifies to: \[ 72u^0v^{11} \]

Step 4: Simplify the Expression

Since any term raised to the power of zero except zero itself is 1, \( u^0 = 1 \). The expression simplifies to: \[ 72v^{11} \]

Step 5: Write Final Expression

The simplified expression, with any negative exponents eliminated, is: \[ 72v^{11} \]

Key Concepts

Multiplying Exponents

When multiplying exponents with the same base, you simply add the exponents together. This rule holds regardless of whether the exponents are positive or negative. In the exercise given, you are presented with the expression \[ (2u^2v^3)^3(3u^{-3}v)^2 \]First, each term needs to be expanded by raising both the coefficients and the variables to the respective powers. For example, in the term \((2u^2v^3)^3\), we raise 2, \(u^2\), and \(v^3\) to the power of 3:- \(2^3 = 8\)- \((u^2)^3 = u^{2 \times 3} = u^6\)- \((v^3)^3 = v^{3 \times 3} = v^9\)Following a similar process for the second expression, \((3u^{-3}v)^2\), yields:- \(3^2 = 9\)- \((u^{-3})^2 = u^{-3 \times 2} = u^{-6}\)- \((v)^2 = v^{2}\)When these expanded expressions are multiplied, the exponents of like bases are added:- \(u^{6} \times u^{-6} = u^{0}\)- \(v^{9} \times v^{2} = v^{11}\)The numbers \(8\) and \(9\) are simply multiplied as usual to get \(72\). The final result is \(72v^{11}\). This demonstrates the power of adding exponents with like bases during multiplication.

Negative Exponents

Negative exponents can seem tricky at first but are actually quite simple! A negative exponent indicates division rather than multiplication. Specifically, it tells us to take the reciprocal of the base and then raise that result to the corresponding positive exponent.For instance, \( u^{-3} \) is interpreted as \( \frac{1}{u^3} \). This is demonstrated in the exercise when expanding \((3u^{-3}v)^2\), yielding \(u^{-6}\). In this exercise, by combining this with \(u^{6}\) from the first expression, the negative exponent is effectively "cancelled out" because \(u^{-6} \times u^{6} = u^{0} = 1\).Using these rules, terms with negative exponents can be moved to the opposite side of a fraction to become positive. It’s a useful technique to simplify and solve problems while ensuring no negative exponents remain in the final expression.

Simplifying Algebraic Expressions

Simplifying algebraic expressions involves combining like terms and ensuring that the expression is as reduced as possible. This means we want to eliminate any unnecessary complex terms, such as negative exponents, and ensure the expression is presented in its simplest form.In the exercise, after multiplying the two expanded expressions, we observe the term \(u^0\), which equals 1. Recognizing and eliminating such terms is a critical part of simplifying, as they do not affect the overall expression.Here’s a tip:- **Look for opportunities to combine terms:** Add or subtract exponents with the same base when multiplying or dividing.- **Eliminate zero exponents:** Remember that any base (except zero) raised to the zero power is 1- **Address negatives effectively:** Use the rules for negative exponents to transform them into positive ones.When these strategies are applied, as in the given problem, the complicated initial expression smoothly simplifies into \(72v^{11}\). This elegant approach highlights the importance of these algebraic principles, making problems more approachable and solvable.