Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(3^{x-1}=28\)

Short answer

\(x \approx 3.634\)

Step-by-step solution

Take the Logarithm on Both Sides

Since the base of the exponential equation is 3, taking the natural logarithm (base e) or base 10 logarithm requires applying the change of base formula later. So, opt for the base 3 logarithm for convenience. As a result, the equation will become \(\log_{3}{3^{x-1}} = \log_{3}{28}\).

Simplify the Equation

Since the base of the logarithm and the base of the exponent on the left-hand side are the same, the left-hand side simplifies directly to the exponent \(x-1\). The equation thus becomes \(x-1 = \log_{3}{28}\).

Solve for the Variable x

We can solve for \(x\) by adding 1 to both sides of the equation, yielding \(x = \log_{3}{28} + 1\). To approximate the result to three decimal places, use a calculator to evaluate the base 3 logarithm, then add 1. The base 3 logarithm can be calculated using the change of base formula if the calculator supports only the natural or base 10 logarithm.

Key concepts

Logarithms

Logarithms are incredibly useful when you are dealing with exponential equations. In simple terms, a logarithm answers the question: "To what power must a certain base be raised, in order to obtain a specific number?" For example, in the expression \( \log_{3}{28} \), it asks, "To what power must 3 be raised to get 28?" Logarithms are essentially the inverse operation of exponentiation.

When working with exponentials, we often take the logarithm on both sides to simplify the process and solve for the unknown variable. In the context of the given exercise, taking \( \log_{3} \) on both sides helps in reducing the complexity of the equation from an exponential form to a linear form, making it easier to solve.

Change of Base Formula

The Change of Base Formula is a handy tool when you need to calculate logarithms with bases other than 10 or \( e \) (the natural logarithm). Many calculators are limited to these bases, so transforming a logarithm into one of these common bases can be crucial.

Here's how the formula works: if you want to compute \( \log_{b}{a} \) using a calculator that only has \( \log_{10} \) and \( \ln \), you use the Change of Base Formula:

• \( \log_{b}{a} = \frac{\log_{c}{a}}{\log_{c}{b}} \)

where \( c \) could be 10 or \( e \). In the original problem, if we choose \( c = 10 \), we can express \( \log_{3}{28} \) as \( \frac{\log_{10}{28}}{\log_{10}{3}} \). This enabled us to calculate base 3 logarithms by using a standard calculator.

Solving Equations

Solving equations is all about finding the value for the variable that satisfies the condition of the equation. In the context of exponential equations, taking logarithms enables us to handle equations with the unknown variable in the exponent.

For the provided exercise, after taking the logarithm on both sides, the equation simplifies to a more familiar linear equation: \( x - 1 = \log_{3}{28} \). Here, solving for \( x \) becomes straightforward:

• Add 1 to both sides to isolate \( x \).
• Calculate \( \log_{3}{28} \) using available logarithm functions and add 1.

Remember, solving equations can involve various steps, but utilizing logarithms as a strategy reveals the path forward, transforming a challenging task into a manageable one. Approximations are often necessary, especially when dealing with irrational numbers, leading to the final decimal form of the solution.