Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(8\left(10^{3 x}\right)=12\)

Short answer

The solution to the equation is \(x = \frac{\ln{1.5}}{3 \ln{10}}\). When this is approximated to three decimal places, we get \(x = -0.369\).

Step-by-step solution

Rewrite the equation

First, the equation \(8\left(10^{3 x}\right)=12\) needs to be rewritten by isolating the exponential expression on one side. This can be done by dividing both sides of the equation by 8, resulting in \(10^{3 x} = \frac{12}{8}\) or \(10^{3 x} = 1.5\).

Apply natural logarithm (ln)

Next, just apply the natural logarithm to both sides of the equation to remove the exponential expression. So, we have \(\ln{\left(10^{3x}\right)} = \ln{1.5}\).

Use properties of Logarithms

Using the property of logarithms, whereby the logarithm of an exponent can be rewritten as the exponent times the logarithm of the base, the equation becomes \(3x \cdot \ln{10} = \ln{1.5}\).

Solve for x

The variable \(x\) can now be isolated by dividing both sides of the equation by \(3 \ln{10}\), giving \(x = \frac{\ln{1.5}}{3 \ln{10}}\).

Key concepts

Natural Logarithm

Natural logarithms are a powerful tool when solving exponential equations, especially when the base of the exponential is not the mathematical constant "e". The natural logarithm, denoted as \( \ln \), uses the base \( e \), which is approximately 2.718. When we apply a natural logarithm to both sides of an equation involving an exponential term, we help simplify the expression by taking advantage of logarithmic properties.

In the equation \( 10^{3x} = 1.5 \), taking the natural logarithm of both sides converts the exponential into something more manageable: \( \ln(10^{3x}) = \ln(1.5) \). This step is essential as it transforms multiplication operations into addition and helps in isolating the variable. Understanding how and when to apply natural logarithms can make solving exponential equations much more straightforward.

Properties of Logarithms

Logarithms have several useful properties that simplify complex calculations. One of the most impactful properties when solving exponential equations is the power rule. This rule states that the logarithm of a number raised to an exponent can be expressed as a product: \( \log_b{(a^n)} = n \cdot \log_b{a} \).

In our problem, applying the natural logarithm and using the power rule helps rewrite \( \ln(10^{3x}) \) as \( 3x \cdot \ln(10) \). This transformation allows us to extract the variable \( x \) from the exponent position and solve it more directly.

• The product rule: \( \log_b{(MN)} = \log_b{M} + \log_b{N} \)
• The quotient rule: \( \log_b{\left(\frac{M}{N}\right)} = \log_b{M} - \log_b{N} \)
• The change of base formula: \( \log_b{a} = \frac{\log_k{a}}{\log_k{b}} \)

Each of these properties is vital in simplifying, expanding, or changing the base of logarithmic functions. By mastering these properties, you can tackle more challenging exponential and logarithmic equations with confidence.

Solving Algebraically

When solving exponential equations algebraically, the main goal is to isolate the variable of interest. This often involves several key steps:

1. Isolating the exponential expression: This was the first step in our given solution. By dividing both sides of the equation by 8, we form \( 10^{3x} = 1.5 \).
2. Applying logarithms: Using the natural logarithm to both sides allows us to harness logarithmic properties to simplify the equation. With \( \ln(10^{3x}) = \ln(1.5) \), we can easily apply the power rule.
3. Simplifying using logarithmic properties: The property \( \ln(10^{3x}) = 3x \cdot \ln(10) \) helps reduce the complexity of the equation and remove the variable from the exponent.

Finally, to solve for the variable \( x \), we rearrange the equation to get \( x = \frac{\ln(1.5)}{3 \cdot \ln{10}} \). Calculating this gives the solution rounded to three decimal places.
By following these systematic steps, we can handle exponential equations effectively, ensuring that each step logically follows to find the solution.