Problem 34
Question
Solve each equation. If necessary, round to the nearest thousandth. $$ 4^{x+1}=28 $$
Step-by-Step Solution
Verified Answer
The solution of the exponential equation \(4^{x+1}=28\) is approximately \(x= 2.169\).
1Step 1: Write the equation
The equation given is \(4^{x+1}=28\). The value of \(x\) needs to be determined.
2Step 2: Taking Logarithms on both sides
In order to get rid of the exponent, and isolate x, apply logarithm on both sides of the equation. Specifically, apply the natural logarithm (ln). This gives us \(\ln{(4^{x+1})} = \ln{28}\). According to the property of logarithms, we can bring \(x+1\) in front. So it simplifies to: \((x+1)\ln4 = \ln28\).
3Step 3: Isolate x
To isolate \(x\), first divide each side by \(\ln(4)\) to get \(x+1 = \frac{\ln28}{\ln4}\). Then subtract 1 from both sides to solve for \(x\). This results in: \(x = \frac{\ln28}{\ln4} - 1\).
4Step 4: Solve using a calculator
Now we use a calculator to evaluate \(\frac{\ln28}{\ln4} - 1\) to the nearest thousandth.
Key Concepts
ExponentiationNatural LogarithmsEquation SolvingMathematical Properties
Exponentiation
Exponentiation is a key operation in mathematics that involves raising a number, known as the base, to the power of an exponent. In the equation, \(4^{x+1}=28\), 4 is the base, and \(x + 1\) is the exponent. This means multiplying 4 by itself \(x + 1\) times.
Exponentiation has some fundamental rules:
Exponentiation has some fundamental rules:
- If the exponent is zero, any non-zero number raised to the power of zero equals 1.
- A base raised to a negative exponent can be rewritten as the reciprocal of the base raised to the positive exponent \(a^{-n} = \frac{1}{a^n}\).
- The product rule states that \(a^m \times a^n = a^{m+n}\).
Natural Logarithms
Natural logarithms, denoted as \(\ln\), are logarithms with the base \(e\), where \(e\) is an irrational number approximately equal to 2.71828. They're particularly useful in calculus and continuous growth problems.
In our given problem, we use natural logarithms to "bring down" the exponent, making the equation more manageable to solve. The equation \(4^{x+1} = 28\) becomes \(\ln{(4^{x+1})} = \ln{28}\). By using the property of logarithms \(\log_b{(a^c)} = c \cdot \log_b{a}\), we transform it to \((x+1)\ln{4} = \ln{28}\).
This transformation simplifies the equation, allowing us to isolate and solve for \(x\). The natural log is preferred here over other logarithms because it is well-suited to continuous exponential growth and decay scenarios.
In our given problem, we use natural logarithms to "bring down" the exponent, making the equation more manageable to solve. The equation \(4^{x+1} = 28\) becomes \(\ln{(4^{x+1})} = \ln{28}\). By using the property of logarithms \(\log_b{(a^c)} = c \cdot \log_b{a}\), we transform it to \((x+1)\ln{4} = \ln{28}\).
This transformation simplifies the equation, allowing us to isolate and solve for \(x\). The natural log is preferred here over other logarithms because it is well-suited to continuous exponential growth and decay scenarios.
Equation Solving
Solving equations is a fundamental skill in mathematics that involves finding the value of the unknown variable that satisfies the equation. In our exercise, we started with an exponential equation \(4^{x+1} = 28\).
To solve for \(x\), we apply the method of taking the natural logarithm of both sides to handle the exponential part. This results in an equation we can manipulate using algebra, \((x+1)\ln{4} = \ln{28}\).
To isolate \(x\), we divide both sides by \(\ln{4}\) and subtract 1, resulting in \(x = \frac{\ln{28}}{\ln{4}} - 1\). This demonstrates the importance of applying the correct mathematical operations and properties in a clear sequence to find the solution.
Using a calculator, we evaluate this expression to find the value of \(x\) to the nearest thousandth. Equation solving often combines multiple concepts in math, highlighting the interconnectedness of these concepts.
To solve for \(x\), we apply the method of taking the natural logarithm of both sides to handle the exponential part. This results in an equation we can manipulate using algebra, \((x+1)\ln{4} = \ln{28}\).
To isolate \(x\), we divide both sides by \(\ln{4}\) and subtract 1, resulting in \(x = \frac{\ln{28}}{\ln{4}} - 1\). This demonstrates the importance of applying the correct mathematical operations and properties in a clear sequence to find the solution.
Using a calculator, we evaluate this expression to find the value of \(x\) to the nearest thousandth. Equation solving often combines multiple concepts in math, highlighting the interconnectedness of these concepts.
Mathematical Properties
Mathematical properties are rules that apply universally to numbers and operations, facilitating the process of equation manipulation and solving.
In our example, several properties come into play:
In our example, several properties come into play:
- Logarithm Properties: Convert multiplication into addition or a power into multiplication, like \(\log_b{(a^c)} = c \cdot \log_b{a}\).
- Exponent Rules: Combine or simplify expressions with the same bases. These include the power rule and product rule outlined in the exponentiation section.
- Algebraic Operations: Allow for isolating variables through operations like addition, subtraction, multiplication, or division.
Other exercises in this chapter
Problem 34
Evaluate \(_{n} C_{x} p^{x} q^{n-x}\) for \(n=7, x=4, p=0.2,\) and \(q=0.8 .\) Round your answer to the nearest thousandth.
View solution Problem 34
Graph each equation. $$ y^{2}-x-2 y+1=0 $$
View solution Problem 35
Distribution \(A\) has 50 data values with mean 40 and standard deviation 2.4 . Distribution \(B\) has 30 data values with mean 40 and standard deviation 2.8 .
View solution Problem 35
Make a box-and-whisker plot for each set of values. $$ 25,25,30,35,45,45,50,55,60,60 $$
View solution