Question

Evaluate the following expressions for \(x=2, y=-3,\) and \(z=-5\) $$ \left(\frac{y}{z}\right)^{x} $$

Short answer

Question: Evaluate the given expression \(\left(\frac{y}{z}\right)^{x}\) for \(x = 2\), \(y = -3\), and \(z = -5\). Answer: \(\frac{9}{25}\)

Step-by-step solution

Substitute given values

First, we will substitute the given values of \(x\), \(y\), and \(z\) into the given expression. Let's plug in \(x = 2\), \(y = -3\), and \(z = -5\). This gives us the expression: \(\left(\frac{-3}{-5}\right)^{2}\)

Simplify the Fraction

Now, we will simplify the fraction. Since both the numerator and the denominator are negative, the fraction becomes positive. Therefore, the fraction simplifies to: \(\left(\frac{3}{5}\right)^{2}\)

Evaluate the Exponent

Finally, we will evaluate the expression by carrying out the power operation. In this case, we have \(\left(\frac{3}{5}\right)^{2}\). We will evaluate this expression by squaring both the numerator and the denominator. That is, we will calculate \((3^2)/(5^2)\). $$ \left(\frac{3}{5}\right)^{2} = \frac{3^2}{5^2} = \frac{9}{25} $$ So, the evaluated expression for the given values of \(x\), \(y\), and \(z\) is \(\frac{9}{25}\).

Key concepts

Exponentiation

Exponentiation is a mathematical operation involving two numbers, the base and the exponent. The base is the number being multiplied, while the exponent indicates how many times the base is multiplied by itself. In the expression \(\left(\frac{-3}{-5}\right)^{2}\), the base is the fraction \(\frac{-3}{-5}\), and the exponent is 2. This means you need to multiply the base by itself twice.

• The operation \(a^b\) where \(a\) is the base and \(b\) is the exponent, translates to multiplying \(a\) by itself \(b\) times.
• For example, \(5^3 = 5 \times 5 \times 5 = 125\).

Exponentiation is not limited to whole numbers—it can also be applied to fractions. By performing \(\left(\frac{3}{5}\right)^2\), each part of the fraction—both the numerator and denominator—is squared. Understanding how to handle exponents is crucial for tackling complex mathematical operations involving powers.

Fraction Simplification

Fraction simplification involves reducing a fraction to its simplest form so that the numerator and denominator have no common factors other than one. Simplifying fractions is a critical step when dealing with algebraic expressions, especially when they involve variables and operations like exponentiation.In the expression \(\left(\frac{-3}{-5}\right)\), the simplification happens because both the numerator (-3) and the denominator (-5) are negative. When both parts of a fraction are negative, they cancel each other out, turning the fraction positive. Thus, the expression \(\left(\frac{-3}{-5}\right)^2\) becomes \(\left(\frac{3}{5}\right)^2\) after simplification.

• A fraction \(\frac{a}{b}\), where both \(a\) and \(b\) are negative, simplifies to \(\frac{|a|}{|b|}\), where \(|\cdot|\) denotes the absolute value.
• Finding the greatest common divisor (GCD) of the numerator and denominator can also help simplify a fraction to its lowest terms.

This simplification is essential before performing any further operations such as exponentiation to ensure accuracy and ease in calculations.

Substitution Method

The substitution method involves replacing variables in an algebraic expression with specific values. This technique is especially useful for evaluating expressions when certain values for the variables are provided.In the given problem, the values of \(x = 2\), \(y = -3\), and \(z = -5\) are substituted into the expression \(\left(\frac{y}{z}\right)^{x}\) to transform it into \(\left(\frac{-3}{-5}\right)^{2}\).

• First, identify the variables in the expression.
• Replace each variable with its corresponding given value.
• Solve the numerical expression that results from the substitution.

This approach allows you to simplify and evaluate complex expressions effectively. Substitution is also widely used in solving equations where multiple variables are involved. By mastering substitution, you can tackle various algebraic problems with greater efficiency.