Problem 19 - Calculus

Question

Solve for \(t .\) Assume \(a\) and \(b\) are positive constants and \(k\) is nonzero. $$a=b^{t}$$

Step-by-Step Solution

Answer

\(t = \frac{\ln(a)}{\ln(b)}\)

Step 1: Introduction to the problem

We need to solve the equation \( a = b^t \) for \( t \). Given that \( a \) and \( b \) are positive constants and \( k \) is nonzero, let's find \( t \).

Step 2: Isolating the exponential term

The equation is \( a = b^t \). Our goal is to solve for \( t \), and in order to do so, we'll need to use logarithmic properties.

Step 3: Apply logarithm to both sides

Take the natural logarithm of both sides to remove the exponent on \( b \). This becomes \( \ln(a) = \ln(b^t) \).

Step 4: Using logarithmic identity

Use the property of logarithms \( \ln(b^t) = t \cdot \ln(b) \) to rewrite the equation \( \ln(a) = t \cdot \ln(b) \).

Step 5: Solving for \( t \)

To isolate \( t \), divide both sides of the equation by \( \ln(b) \). This gives \( t = \frac{\ln(a)}{\ln(b)} \).

Key Concepts

Natural Logarithms

Natural logarithms are vital in solving equations involving exponential forms. Such logarithms use the constant base, denoted by the letter "e," which approximates 2.71828. The natural logarithm of a number is abbreviated as \( \ln \), and it is the power to which \( e \) must be raised to obtain that number. For instance, if \( x = e^y \), then \( \ln(x) = y \). This function is helpful for reversing exponential functions, turning multiplication into addition, which simplifies solving for unknowns in equations.
In our exercise, applying the natural logarithm to both sides of the equation \( a = b^t \) gives \( \ln(a) = \ln(b^t) \). This conversion is crucial because it allows us to handle the exponent, using logarithmic properties to simplify and solve the equation efficiently.

Logarithmic Properties

Understanding logarithmic properties is key to transforming and solving exponential equations. Logarithms can simplify multiplication into addition, division into subtraction, and powers into multiplication. Here are some fundamental properties:

• \( \ln(1) = 0 \): Because any number raised to the power of 0 equals 1.
• \( \ln(x^y) = y \cdot \ln(x) \): When an argument is an exponent, it can be brought down as a coefficient.
• \( \ln(x \cdot y) = \ln(x) + \ln(y) \): Logarithm of a product is the sum of the logarithms.

These properties allow us to manipulate expressions to solve for unknown variables efficiently. In the exercise, we use \( \ln(b^t) = t \cdot \ln(b) \), a critical step due to the logarithmic power property, turning the exponential problem into a linear equation in terms of \( t \).

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to isolate the variable of interest. When dealing with logarithms, it often includes applying logarithmic properties and basic algebraic techniques to solve for the unknown. In the equation from the exercise, \( \ln(a) = t \cdot \ln(b) \), our goal is to isolate \( t \).
By dividing both sides by \( \ln(b) \), we effectively solve for \( t \): \( t = \frac{\ln(a)}{\ln(b)} \). This division is a simple yet powerful algebraic technique that transforms the expression, isolating \( t \) on one side of the equation. Understanding these fundamental steps in algebraic manipulation is essential for solving logarithmic equations and can be broadly applied to various mathematical problems.