Question

If \(f(x)=\log _{a} x,\) show that \(f\left(x^{\alpha}\right)=\alpha f(x)\)

Short answer

Since \( \log_a (x^{\alpha}) = \alpha \log_a x \), it follows that \( f(x^{\alpha}) = \alpha f(x) \).

Step-by-step solution

- Understand the Function

The function given is a logarithmic function: \( f(x) = \log_a x \). The goal is to show that applying this function to \( x^{\alpha} \) yields \( \alpha f(x) \).

- Substitute in the Function

Apply the function to \( x^{\alpha} \). This means we need to substitute \( x^{\alpha} \) into the original function: \( f(x^{\alpha}) = \log_a (x^{\alpha}) \).

- Apply Logarithm Property

Use the logarithm property that states \( \log_a (b^c) = c \log_a b \). Therefore, \( f(x^{\alpha}) = \log_a (x^{\alpha}) = \alpha \log_a x \).

- Relate to Original Function

Recognize that \( \log_a x = f(x) \). Thus, \( f(x^{\alpha}) = \alpha f(x) \).

- Conclude the Identity

The final expression \( f(x^{\alpha}) = \alpha f(x) \) confirms the desired result using the properties of logarithms.

Key concepts

Logarithm Properties

Logarithmic functions have several key properties that are really useful in solving equations, especially ones like this exercise. The property we used here is crucial: \[ \log_a(b^c) = c \log_a(b) \].
This means when you take the logarithm of a number raised to a power, you can 'pull' the exponent in front as a multiplier.
Another essential property is the change of base formula: \[ \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \].
This helps when you need to calculate logarithms in different bases.

These properties simplify complex logarithmic expressions and make it easier to solve equations. For instance, logarithms turn multiplication into addition and division into subtraction:

• Multiplication: \[ \log_a(xy) = \log_a(x) + \log_a(y) \]
• Division: \[ \log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y) \]

Understanding these properties is vital for handling logarithmic functions confidently.

Exponential Functions

Exponential functions are the flip side of logarithms. The function \[ f(x) = a^x \] describes how numbers grow or decay exponentially.
Here's a handy connection: \[ \log_a(a^x) = x \] and \[ a^{\log_a(x)} = x \].
These identities show the inverse relationship between exponential and logarithmic functions.

Exponential functions grow very quickly. For example, if you have \[ 2^x \], as x increases, the value of the function shoots up.
This rapid growth is why exponential functions are used to model things like population growth or radioactive decay.
On the other hand, logarithmic functions grow very slowly. If you compare \[ \log_2(x) \] to \[ 2^x \], you'll notice that \[ \log_2(x) \] increases at a much slower pace.

Understanding the behavior of exponential functions helps to interpret the effects when they are involved in logarithmic expressions, like in our exercise.

Function Substitution

Function substitution is a method used to simplify complex expressions by replacing a part of the function with another value or function.
In the given exercise, we substituted \[ x^{\alpha} \] into the function \[ f(x) = \log_a(x) \].
So, we looked at \[ f(x^{\alpha}) \] and replaced every \[ x \] with \[ x^{\alpha} \].

This gives us \[ f(x^{\alpha}) = \log_a(x^{\alpha}) \].
Knowing the property of logarithms \[ \log_a(b^c) = c \log_a(b) \], we transformed this into \[ \alpha \log_a(x) \]. By recognizing that \[ \log_a(x) \] is just \[ f(x) \], it simplified to \[ \alpha f(x) \].

Substitution helps tackle intricate parts of a problem by breaking them down into simpler, familiar components.
This method is widely used in calculus and higher-level mathematics to make solving equations more manageable.
It's also a powerful tool for understanding and demonstrating proofs, just like in this exercise.