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Chapter 3: Problem 6

In Exercises 5 - 12, determine whether each \( x \)-value is a solution (or an approximate solution) of the equation. \( 2^{3x + 1} = 32 \) (a) \( x = -1 \) (b) \( x = 2 \)

Short Answer

Expert verified
Neither \(x = -1\) nor \(x = 2\) is a solution to the equation \(2^{3x + 1} = 32\).

Step by step solution

01

Checking Solution for \(x = -1\)

Substitute \(x = -1\) into the equation. This gives us \(2^{3*(-1) + 1} = 32\), which simplifies to \(2^{-2} = 32\). Further breaking it down we get \(1/4 = 32\). This is not true, hence \(x = -1\) is not a solution to the equation.
02

Checking Solution for \(x = 2\)

Substitute \(x = 2\) into the equation. This gives us \(2^{3*2 + 1} = 32\), which simplifies to \(2^{7} = 32\). We can rewrite 32 as \(2^5\), so if both sides of the equation are equal, then they should simplify to the same power of 2. Because 7 is not equal to 5, \(x = 2\) is not a solution to the equation.

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Key Concepts

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

Solution Verification
Verifying solutions for equations is an important skill that helpsensure our answers are correct. One way to verify solutions is by substituting the potential solution back into the original equation to see if it makes the equation true. Let's use the equation \(2^{3x + 1} = 32\) as an example.
  • Substitute the suggested \(x\)-value into the equation.
  • Simplify both sides of the equation.
  • Check whether both sides are equal.
For instance, in our exercise, we try values \(x = -1\) and \(x = 2\).If when substituting either of these into the equation, both sides become equal, that means the \(x\)-value is indeed a solution.However, if after substitution the equation does not hold, then that \(x\)-value is not a solution.
Exponential Functions
Exponential functions are functions where the variable appears in the exponent.They are written in the form \(f(x) = a^{bx+c}\), where \(a\) is a constant, and \(b\) and \(c\) are also constants.In these functions, even slight changes in the variable \(x\) cause significant changes in the value of the function.
In the equation \(2^{3x + 1} = 32\), consider the exponential part \(2^{3x + 1}\). The base here is \(2\).The power, or exponent, changes depending on the value of \(x\).
  • Exponential growth happens when the base is greater than 1.
  • Exponential decay occurs when the base is between 0 and 1.
Exponential functions powerfully model many real-life situations like population growth and radioactive decay.
Power of a Number
The power of a number tells us how many times to multiply the number by itself.For instance, the power in \(2^{3x + 1}\) represents the exponent that follows the base of 2.Manipulating powers is crucial when simplifying equations like \(2^{3x + 1} = 32\).
For instance, to see if \(2^{7} = 32\), we need to see if both sides have the same power when expressed as powers of 2.
  • \(32\) can be written as \(2^5\).
  • To have equality, we need \(2^{3x + 1}\) to simplify to \(2^5\).
The equation's balance is achieved when the powers are equal, but here \(7eq5\), therefore \(x = 2\) is not the solution.

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Most popular questions from this chapter

Fill in the blanks. To ________ an equation in \( x \) means to find all values of \( x \) for which the equation is true.

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A cup of water at an initial temperature of \( 78^{\circ}C \) is placed in a room at a constant temperature of \( 21^{\circ}C \). The temperature of the water is measured every 5 minutes during a half-hour period.The results are recorded as ordered pairs of the form \( (t \), \( T) \), where \( t \) is the time (in minutes) and \( T \) is the temperature (in degrees Celsius). \( \left(0, 78.0^{\circ}\right) \), \( \left(5 , 66.0^{\circ}\right) \), \( \left(10, 57.5^{\circ}\right) \), \( \left(15 , 51.2^{\circ}\right) \), \( \left(20 , 46.3^{\circ}\right) \), \( \left(25, 42.4^{\circ}\right) \), \( \left(30 , 39.6^{\circ}\right) \) (a) The graph of the model for the data should be asymptotic with the graph of the temperature of the room. Subtract the room temperature from each of the temperatures in the ordered pairs. Use a graphing utility to plot the data points \( \left(t , T\right) \) and \( \left(t, T - 21\right) \). (b) An exponential model for the data \( \left(t, T - 21\right) \) is given by \( T - 21 = 54.4\left(0.964\right)^t \) Solve for \( T \) and graph the model. Compare the result with the plot of the original data. (c) Take the natural logarithms of the revised temperatures. Use a graphing utility to plot the points \( \left(t, In\left(T - 21\right)\right) \) and observe that the points appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. This resulting line has the form \( In\left(T - 21\right) = at + b \). Solve for \( T \), and verify that the result is equivalent to the model in part (b). (d) Fit a rational model to the data. Take the reciprocals of the \( y \)-coordinates of the revised data points to generate the points \( \dfrac{1}{T - 21} = at + b \). Solve for \( T \), and use a graphing utility to graph the rational function and the original data points. (e) Why did taking the logarithms of the temperatures lead to a linear scatter plot? Why did taking the reciprocals of the temperatures lead to a linear scatter plot?
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