Problem 20 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve each exponential equation by taking the logarithm on both sides. Express the solution set in terms of logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$e^{x}=0.83$$

Step-by-Step Solution

Answer

The solution to the equation \(e^{x} = 0.83\) rounded off to two decimal places is \(x = \ln(0.83) = -0.19\).

Step 1: Identify the base and exponent

The given equation is \(e^{x} = 0.83\). The base of the exponent is the natural number \(e\) and the exponent itself is \(x\). The equation is solved by extracting \(x\) from the exponential term.

Step 2: Take the natural logarithm of both sides

This is done to get the variable out of the exponent. The property of logarithms we use here is \(\ln(a^b) = b\ln(a)\). Thus, when we take the natural logarithm on both sides, the equation becomes: \(\ln(e^x) = \ln(0.83)\). Simplify the left side of this equation using one of the logarithm properties: \(\ln(e^x)\) equals \(x*ln(e)\). But since \(\ln(e)\) equals 1, the left side of the equation simplifies to \(x\). So, \(x = \ln(0.83)\).

Step 3: Find the decimal approximation of the solution

This step requires a calculator to get a decimal approximation of the undefined logarithm value - \(\ln(0.83)\). The solution should be rounded off to two decimal places.

Key Concepts

Natural Logarithm

When we discuss the natural logarithm, we are focusing on a logarithm with the base of Euler's number, denoted as "e". This number "e" is approximately equal to 2.71828 and is a fundamental constant in mathematics, particularly useful in calculus.

• The natural logarithm is represented by the symbol \(\ln\).
• It serves to "undo" the exponential function \(e^x\).
• The equation \(\ln(e^x) = x\) is based on the property that the natural logarithm of \(e\) raised to any power is simply that power.

Understanding this property helps us to solve equations that involve \(e\), as it effectively lets us "bring down" the exponent to a more solvable form. In our original problem of solving \(e^x = 0.83\), taking the natural logarithm of both sides allowed us to simplify the equation into something more manageable: \(x = \ln(0.83)\). This step makes it easier to isolate and solve for \(x\).
Natural logarithms provide a straightforward method for dealing with continuous growth or decay models.

Decimal Approximation

Once we've used logarithms to isolate a variable, we often need a decimal approximation of the solution, especially when dealing with irrational numbers. An irrational number is a number that cannot be exactly expressed as a ratio of two integers. Therefore, its decimal form is non-repeating and non-terminating.

• For the problem \(x = \ln(0.83)\), we typically rely on a calculator to obtain the value of \(\ln(0.83)\).
• The result is approximately \(-0.18633\), but to comply with the exercise requirements, we round this number to two decimal places.
• Thus, the final value is typically expressed as \(-0.19\).

Calculators compute these values by utilizing algorithms that approximate the logarithmic functions. Accurate decimal approximations help in providing a clearer understanding and concrete numeric interpretation of our mathematical results. Additionally, rounding to a specific decimal point ensures consistency and readability, particularly in a classroom or academic setting.

Logarithm Properties

Logarithms possess various properties that make solving exponential equations more manageable. One of the most crucial properties utilized in the previous steps is the power rule for logarithms. This key property was instrumental in solving the equation \(e^x = 0.83\).

• Power Rule: \(\ln(a^b) = b\ln(a)\). This rule helps to "bring down" the exponent, simplifying equations and expressions.
• Change of Base Formula: Although not directly used in our exercise, knowing that \(\log_b(a) = \frac{\ln(a)}{\ln(b)}\) can be very useful in converting between different logarithmic bases.
• Product and Quotient Rules: These are: \(\log_b(m \cdot n) = \log_b(m) + \log_b(n)\) and \(\log_b(\frac{m}{n}) = \log_b(m) - \log_b(n)\), respectively. These enable simplification of expressions involving multiplied or divided terms.

Understanding these properties is vital as they can transform and greatly simplify complex expressions into calculable forms. Mastery of these properties enables students to effectively solve a wide range of logarithmic and exponential equations.