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Chapter 5: Problem 45

Exer. 45-46: Sketch the graph of \(f\), and use the change of base formula to approximate the \(x\)-intercept. $$ f(x)=4^{x}-3 $$

Short Answer

Expert verified
The x-intercept of the function \( f(x) = 4^x - 3 \) is approximately 0.7925.

Step by step solution

01

Understand the Function

The function given is \( f(x) = 4^x - 3 \). This is an exponential function where the base is 4. The graph of \( 4^x \) alone is an increasing curve, so \( f(x) \) will be the same curve shifted down by 3 units.
02

Sketch the Graph

To sketch the graph, start with the basic shape of \( y = 4^x \), an exponentially increasing curve passing through \( (0, 1) \). Since our function is \( 4^x - 3 \), shift this entire graph downward by 3 units. The curve will approach \( y = -3 \) as \( x \) approaches negative infinity and will continue to increase past the x-axis.
03

Determine the x-intercept

The x-intercept is the point where the function crosses the x-axis, which means \( f(x) = 0 \). We set the equation to zero to find this point: \[ 4^x - 3 = 0 \] Simplifying gives \[ 4^x = 3 \].
04

Apply the Change of Base Formula

To solve \( 4^x = 3 \), take the logarithm of both sides. You can use any base, but let's use base 10 for easier computation: \[ \log_{10}(4^x) = \log_{10}(3) \] Apply the power rule: \[ x \cdot \log_{10}(4) = \log_{10}(3) \] Solve for \( x \): \[ x = \frac{\log_{10}(3)}{\log_{10}(4)} \]
05

Approximate the x-intercept

Use a calculator to find the values: \( \log_{10}(3) \approx 0.4771 \) and \( \log_{10}(4) \approx 0.6021 \). So, \[ x \approx \frac{0.4771}{0.6021} \approx 0.7925 \]. This is the approximate x-intercept.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Graphing Functions
Graphing functions can seem intimidating at first, but breaking it down can simplify the process. When dealing with exponential functions like \( f(x) = 4^x - 3 \), the graph can be an insightful tool to understand their behavior. Start by identifying the basic graph, which in this case, is \( y = 4^x \), characterized by its rapid increase as \( x \) becomes larger. This function crosses the y-axis at (0,1). However, our function \( f(x) \) is shifted downwards by 3 units because of \(-3\).
  • Visualize the basic graph of \( y = 4^x \).
  • Shift this graph down by 3 units since every point on \( f(x) \) is decreased by 3.
  • Understand that the line will get closer and closer to \( y = -3 \) as \( x \) goes towards negative infinity, but never actually touch it.
  • The graph will continue to rise and eventually cross the x-axis as \( x \) gets larger.
This visual model helps us comprehend how the function behaves across the x-values.
Change of Base Formula
The change of base formula is a useful tool in solving logarithmic equations involving exponential functions. For our given function \( f(x) = 4^x - 3 \), finding the x-intercept involves solving \( 4^x = 3 \). Using the change of base formula can be necessary because this equation doesn't simplify easily within the integers.
  • The formula allows you to change the base of the logarithm from base 4 to your chosen base, usually 10 or \( e \) for convenience.
  • In using base 10, the formula becomes: \( \log_{4}(x) = \frac{\log_{10}(x)}{\log_{10}(4)} \).
  • This transformation makes it straightforward to solve using a calculator, thus providing an accurate approximation.
Mastering this formula equips you with the ability to handle a wide range of logarithmic problems effectively.
X-Intercepts
Finding the x-intercept of an exponential function involves determining when the function equals zero. This intercept is where the graph crosses the x-axis. For \( f(x) = 4^x - 3 \), setting the equation to zero gives us \( 4^x = 3 \).
  • At this point, we are looking for the x-value where the y-value is zero.
  • Simplifying the equation leads to using logarithms because \( x \) must be isolated as \( x = \log_{4}(3) \).
  • Approximating this value with the change of base formula provides a more tangible understanding.
  • After calculation, the approximate x-intercept for this function is \( x \approx 0.7925 \). It tells exactly where on the x-axis our exponential curve passes through.
This conclusion enhances our grasp of the function's behavior at the intercept.
Exponential Equations
Exponential equations like \( 4^x = 3 \) are equations where the exponent is the variable. These equations often involve growth or decay processes and require methods such as logarithms to solve.
  • Recognize the type of exponential growth depicted in \( f(x) = 4^x - 3 \), fast increasing as we progress along the x-axis.
  • Solving the equation \( 4^x = 3 \) involves taking the logarithm of both sides, simplifying the variable's isolation.
  • By using logarithmic properties, specifically the power rule \( x \cdot \log(4) = \log(3) \), the exponent can be calculated.
  • This demonstration of exponential equations reveals patterns within the applications of growth models.
These competencies are crucial for tackling various scientific and financial problems involving exponential relations.

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Most popular questions from this chapter

If \(p\) denotes the selling price (in dollars) of a commodity and \(x\) is the corresponding demand (in number sold per day), then the relationship between \(p\) and \(x\) is sometimes given by \(p=p_{0} e^{-a x}\), where \(p_{0}\) and \(a\) are positive constants. Express \(x\) as a function of \(p\).

A drug is eliminated from the body through urine. Suppose that for a dose of 10 milligrams, the amount \(A(t)\) remaining in the body \(t\) hours later is given by \(A(t)=10(0.8)^{t}\) and that in order for the drug to be effective, at least 2 milligrams must be in the body. (a) Determine when 2 milligrams is left in the body. (b) What is the half-life of the drug?

(a) \(f(x)=\ln (x+1)+e^{x}, \quad x=2\) (b) \(g(x)=\frac{(\log x)^{2}-\log x}{4}, x=3.97\)

Particle velocity A very small spherical particle (on the order of 5 microns in diameter) is projected into still air with an initial velocity of \(v_{0} \mathrm{~m} / \mathrm{sec}\), but its velocity decreases because of drag forces. Its velocity \(t\) seconds later is given by \(v(t)=v_{0} e^{-a t}\) for some \(a>0\), and the distance \(s(t)\) the particle travels is given by $$ s(t)=\frac{v_{0}}{a}\left(1-e^{-a t}\right) . $$ The stopping distance is the total distance traveled by the particle. (a) Find a formula that approximates the stopping distance in terms of \(v_{0}\) and \(a\). (b) Use the formula in part (a) to estimate the stopping distance if \(v_{0}=10 \mathrm{~m} / \mathrm{sec}\) and \(a=8 \times 10^{5}\).

Sketch the graph of \(f\). $$ f(x)=\log _{2}\left(x^{3}\right) $$
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