Question

Consider the following exponential functions: \(\cdot f(x)=4^{x}\) \(\cdot g(x)=\left(\frac{1}{4}\right)^{x}\) \(\cdot h(x)=2^{x}\) a) Which is greatest when \(x=5 ?\) b) Which is greatest when \(x=-5 ?\) c) For which value of \(x\) do all three functions have the same value? What is this value?

Short answer

a) f(x) is greatest when x=5. b) g(x) is greatest when x=-5. c) All functions are equal when x=0.

Step-by-step solution

Compare values at x=5

Evaluate each function at x=5:1. For f(x) = 4^x: \[ f(5) = 4^5 = 1024 \]2. For g(x) = \left( \frac{1}{4} \right)^x: \[ g(5) = \left( \frac{1}{4} \right)^5 = \left( \frac{1}{4} \right)^5 = \frac{1}{1024} \]3. For h(x) = 2^x: \[ h(5) = 2^5 = 32 \] Therefore, at x=5, f(x) is the greatest.

Compare values at x=-5

Evaluate each function at x=-5:1. For f(x) = 4^x: \[ f(-5) = 4^{-5} = \frac{1}{1024} \]2. For g(x) = \left( \frac{1}{4} \right)^x: \[ g(-5) = \left( \frac{1}{4} \right)^{-5} = 4^5 = 1024 \]3. For h(x) = 2^x: \[ h(-5) = 2^{-5} = \frac{1}{32} \]Therefore, at x=-5, g(x) is the greatest.

Determine the value of x where all functions are equal

Set the functions equal to each other:1. Set f(x) = h(x): \[ 4^x = 2^x \] Rewrite 4 as 2^2: \[ (2^2)^x = 2^x \] Simplify: \[ 2^{2x} = 2^x \] Since the bases are the same, set the exponents equal: \[ 2x = x \] Thus, no solution from this equation.2. Set g(x) = h(x): \[ \left( \frac{1}{4} \right)^x = 2^x \] Rewrite \( \frac{1}{4} \) as \( (2^{-2}) \): \[ (2^{-2})^x = 2^x \] Simplify: \[ 2^{-2x} = 2^x \] Set the exponents equal: \[ -2x = x \] Solve for x: \[ -2x = x \] \[ -3x = 0 \] \[ x = 0 \] Therefore, the three functions are equal at x = 0.

Key concepts

Function Comparison

In this exercise, we compare different exponential functions: \( f(x) = 4^x \), \( g(x) = \left( \frac{1}{4} \right)^x \), and \( h(x) = 2^x \). Each function has unique characteristics based on its base.
Exponential functions grow or decay at different rates depending on the value and form of their base. Here, we evaluate each function at specific points in order to directly compare their magnitudes.
At \( x = 5 \):

• \( f(5) = 4^5 = 1024 \)
• \( g(5) = \left( \frac{1}{4} \right)^5 = \frac{1}{1024} \)
• \( h(5) = 2^5 = 32 \)

Clearly, \( f(5) = 1024 \) is the greatest value.
At \( x = -5 \):

• \( f(-5) = 4^{-5} = \frac{1}{1024} \)
• \( g(-5) = \left( \frac{1}{4} \right)^{-5} = 4^5 = 1024 \)
• \( h(-5) = 2^{-5} = \frac{1}{32} \)

In this case, \( g(-5) = 1024 \) is the greatest.
This comparison reveals how different bases in exponential functions lead to vastly different values, especially as \( x \) grows positive or negative.

Solving Exponential Equations

To find where all three functions \( f(x), g(x), \) and \( h(x) \) are equal, we set the functions equal to each other and solve for \( x \).
First, compare \( f(x) \) and \( h(x) \): \[ 4^x = 2^x \] Rewrite \( 4 \) as \( 2^2 \): \[ (2^2)^x = 2^x \] Simplify to: \[ 2^{2x} = 2^x \] As the bases are the same, set the exponents equal: \[ 2x = x \] This reveals no valid solution for \( x \).
Next, compare \( g(x) \) and \( h(x) \): \[ \left( \frac{1}{4} \right)^x = 2^x \] Rewrite \( \frac{1}{4} \) as \( (2^{-2}) \): \[ (2^{-2})^x = 2^x \] Simplify to: \[ 2^{-2x} = 2^x \] Set the exponents equal: \[ -2x = x \] Solve for \( x \): \[ -2x = x \] \[ -3x = 0 \] \[ x = 0 \] Thus, \( x = 0 \) is where all three functions intersect, making their values equal.

Exponent Properties

Understanding the properties of exponents is crucial in solving exponential equations and comparing functions. Here are some essential properties:

• \( a^0 = 1 \) for any non-zero base \( a \)
• \( a^{-n} = \frac{1}{a^n} \)
• \( (a^m)^n = a^{m \times n} \)
• \( a^m \times a^n = a^{m + n} \)
• \( \left( \frac{a}{b} \right)^n = \frac{a^n}{b^n} \)

Consider the base manipulations used in our equations:
- \(4^x = (2^2)^x = 2^{2x} \): Simplifying this equation is key to comparing it with another function.
- \(\left( \frac{1}{4} \right)^x = (2^{-2})^x = 2^{-2x} \): This manipulation helps us equate the function to other bases of 2.
These properties enable solving more complex exponential equations by reducing them to comparables forms. Using these rules can clarify the steps in solving, simplifying, and comparing exponential expressions.