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In Exercises 29 - 70, solve the exponential equation algebraically. Approximate the result to three decimal places. \( 3\left(5^{x - 1}\right) = 21 \)

Short Answer

Expert verified
The solution to the equation is \(x \approx 2.209\).

Step by step solution

01

Isolate the exponential term

To isolate the exponential term on one side of the equation, we'll divide both sides by 3. This will give us: \[5^{x - 1} = \frac{21}{3} = 7\]
02

Apply logarithm on both sides

Since 5 and 7 are both positive numbers and the base of the exponential term is 5, we can apply logarithm of base 5 on both sides. This will result in: \[ \log_{5}(5^{x - 1}) = \log_{5}(7)\] \[ x - 1 = \log_{5}(7)\]
03

Manipulate for the solution

Adding 1 to both sides for the final solution gives us: \[ x = \log_{5}(7) + 1 \]
04

Approximate the solution

Finally, we need to approximate the result to three decimal places. Using a calculator, we get: \[ x = \log_{5}(7) + 1 \approx 2.209 \]

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Key Concepts

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

Logarithms
Logarithms are a fundamental concept in mathematics, especially when dealing with exponential equations. They provide a method to solve equations where the variable is an exponent. Logarithms are essentially the inverse operation of exponentiation. For example, if you have an equation like \(a^x = b\), applying the logarithm of base \(a\) to both sides can help isolate \(x\).

To break it down further:
  • In exponential equations, the variable is in the exponent, making them tricky to solve through simple algebraic means.
  • Using the property of logarithms, \(\log_{a}(a^x) = x\), we can simplify the equation considerably.
In the exercise, applying a base 5 logarithm on both sides of \(5^{x-1} = 7\) simplifies it to \(x-1 = \log_{5}(7)\). This step is crucial as it transforms the equation into a linear form, making it easier to solve for \(x\). The process of using logarithms connects the base with the exponent and the result, allowing a path to solve for unknowns efficiently.
Algebraic Manipulation
Algebraic manipulation is the process of rearranging and simplifying equations to find solutions. In solving the given exponential equation, algebraic manipulation plays a critical role in both isolating terms and simplifying expressions.

Here's the step-by-step process in the exercise:
  • Initially, the equation is \(3(5^{x-1}) = 21\). We begin by dividing both sides by 3 to isolate the exponential term, resulting in \(5^{x-1} = 7\).
  • Once isolated, apply the logarithm to simplify the variable's exponent to a more manageable form.
  • Finally, solve for \(x\) by performing operations like addition or subtraction, enhancing the simplicity of the equation.
The key is understanding how to manipulate terms while maintaining equality, ensuring the original equation remains true. This might involve operations like addition, subtraction, multiplication, or division. Mastery of these operations is essential, providing a solid foundation for further mathematical advancements.
Approximation Methods
Approximation methods are essential when an exact analytical solution is complex or impossible. After simplifying and solving the algebraic form of an exponential equation, there's often a need to find practical numerical values. In the exercise provided, the aim is to approximate the result to three decimal places.

Several tools and techniques can assist with approximation:
  • Scientific calculators or technology like computer algebra systems can compute logarithm values accurately.
  • Rounding to a specified number of decimal places ensures that the solution is practical for real-world applications.
In this case, using the calculator to find \(x = \log_{5}(7) + 1\) provides approximately 2.209 when rounded to three decimal places. Approximation makes complex numbers more manageable, ensuring that results are usable and understandable in various contexts.

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Most popular questions from this chapter

In Exercises 81 - 112, solve the logarithmic equation algebraically. Approximate the result to three decimal places. \( 2 \ln x = 7 \)

The \( IQ \) scores for a sample of a class of returning adult students at a small northeastern college roughly follow the normal distribution \( y = 0.0266e^{-(x - 100)^2/450} \) , \( 70 \le x \le 115 \), where \( x \) is the \( IQ \) score. (a) Use a graphing utility to graph the function. (b) From the graph in part (a), estimate the average \( IQ \)score of an adult student.

In Exercises 81 - 112, solve the logarithmic equation algebraically. Approximate the result to three decimal places. \( \ln x + \ln\left(x + 1\right) = 1 \)

In Exercises 121 - 128, solve the equation algebraically. Round the result to three decimal places. Verify your answer using a graphing utility. \( \dfrac{1 - \ln x}{x^2} = 0 \)

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