Problem 500 - Precalculus

Question

Rewrite \(a^{-\frac{2}{5}}=b\) as an equivalent logarithmic equation.

Step-by-Step Solution

Answer

\(\log_a b = -\frac{2}{5}\)

Step 1: Understand the Given Equation

The equation provided is an exponential equation: \(a^{-\frac{2}{5}}=b\). Our goal is to convert this exponential equation into its equivalent logarithmic form.

Step 2: Recall the Exponential-Logarithm Relationship

The relationship between exponents and logarithms is given by: if \(x^y = z\), then it can be rewritten in logarithmic form as \(\log_x z = y\).

Step 3: Identify Corresponding Terms

By comparing \(a^{-\frac{2}{5}}=b\) to \(x^y = z\), we identify that \(x\) corresponds to \(a\), \(y\) corresponds to \(-\frac{2}{5}\), and \(z\) corresponds to \(b\).

Step 4: Rewriting in Logarithmic Form

Utilizing the identified terms and the relationship, we rewrite the equation in logarithmic form: \(\log_a b = -\frac{2}{5}\).

Key Concepts

Exponentiation

Exponentiation is a mathematical operation involving a base and an exponent. It is written in the form \(a^b\), where \(a\) is the base, and \(b\) is the exponent. This operation is equivalent to multiplying the base \(b\) times. For instance, \(a^3\) represents \(a \times a \times a\). It is important to understand how negative and fractional exponents work.

• Negative Exponents: A negative exponent means the reciprocal of the base raised to the absolute value of the exponent. For example, \(a^{-1} = \frac{1}{a}\) and \(a^{-2} = \frac{1}{a^2}\).
• Fractional Exponents: A fractional exponent, like \(a^{\frac{1}{2}}\), represents the root of the base. Specifically, \(a^{\frac{1}{n}}\) is the \(n\)-th root of \(a\).

For the given problem, \(a^{-\frac{2}{5}}\) represents the reciprocal of the fifth root of \(a\) squared. To visualize this, think of it as finding the fifth root of \(a\), squaring it, then taking the reciprocal of that result.

Logarithmic Form Conversion

Converting from exponential form to logarithmic form is a common task in algebra. It involves switching between two mathematical expressions that describe the same relationship. To convert an equation in the form of \(a^b = c\) to its logarithmic form, we use the relationship \(\log_a c = b\). The base \(a\) in the exponential form becomes the base of the logarithm, \(c\) becomes the argument of the logarithm, and the exponent \(b\) is the result of the logarithm. For example, if we start with an equation like \(a^{-\frac{2}{5}} = b\), we need to identify:

• Base \(a\): This stays the same.
• Exponent \(-\frac{2}{5}\): This becomes the result of the logarithm.
• Result \(b\): This is what the logarithm evaluates.

The logarithmic form would be \(\log_a b = -\frac{2}{5}\). In practical terms, this expression tells you the power you raise \(a\) to in order to get \(b\).

Exponential-Logarithm Relationship

Understanding the exponential-logarithm relationship is crucial for solving equations involving exponentials and logarithms. This relationship provides a bridge between the two forms:

• Exponential Form: \(x^y = z\)
• Logarithmic Form: \(\log_x z = y\)

The exponential form indicates that a base \(x\) raised to an exponent \(y\) results in \(z\). Conversely, the logarithmic form tells you that \(y\) is the exponent to which \(x\) must be raised to get \(z\).In the given example \(a^{-\frac{2}{5}} = b\), the transformation to logarithmic form \(\log_a b = -\frac{2}{5}\) utilizes this fundamental relationship. This means that \(-\frac{2}{5}\) is the exponent that \(a\) must achieve to result in \(b\). This conversion is helpful because logarithms can simplify solving equations, particularly those not easily managed by simple algebraic techniques.Keep in mind that logarithms and exponentials are inverse functions, meaning they can undo each other. This property is useful for many mathematical operations and problem-solving scenarios.