Problem 32 - Calculus of a Single Variable

Question

Solving an Equation In Exercises \(25-34,\) solve the equation accurate to three decimal places. $$ \log _{10}(t-3)=2.6 $$

Step-by-Step Solution

Answer

The solution for \( t \) is approximately 403.776.

Step 1: Rewrite the logarithmic equation in exponential form

First, remember that a logarithm equation \( \log _{b}a=c \) can be rewritten in exponential form as \( b^c = a \). Therefore, the equation \( \log_{10}(t-3) = 2.6 \) can be rewritten as \( 10^{2.6} = t - 3 \)

Step 2: Solve for the variable t

Once the equation has been rewritten in its exponential form, the variable \( t \) can be solved by moving \( 3 \) from the right side to the left side of the equation. It will result in to the equation: \( t = 10^{2.6} + 3 \)

Step 3: Find t accurate to three decimal places

Finally, calculate the value of \( t \) using a calculator and make sure it is accurate to three decimal places. This calculation will provide the solution for \( t \)

Key Concepts

Exponential Form

A logarithmic equation can be turned into exponential form, which often simplifies solving for variables. In a logarithmic equation like \( \log_{b}(a) = c \), the equivalent exponential form is \( b^c = a \). Here, \( b \) is the base of the logarithm, \( a \) is the antilogarithm or result, and \( c \) is the logarithm. This conversion is beneficial because exponential equations can make it clearer how to isolate the variable you're trying to find.
In our original problem, \( \log _{10}(t-3)=2.6 \) is rewritten in exponential form as \( 10^{2.6} = t - 3 \). This gives a straightforward equation to work with.

• Logarithmic form: \( \log_{b}(a) = c \)
• Exponential form: \( b^c = a \)

Converting between these two forms is a fundamental skill for solving logarithmic equations effectively.

Solve for a Variable

To solve for a variable in an exponential equation, the process involves isolating the variable on one side of the equation. This might involve simple arithmetic operations such as addition, subtraction, multiplication, or division.
For example, in the equation \( 10^{2.6} = t - 3 \), the variable \( t \) can be isolated by adding 3 to both sides. This results in \( t = 10^{2.6} + 3 \). Now, the variable \( t \) stands alone and can easily be evaluated using a calculator.
Key steps include:

• Isolate the variable of interest.
• Use basic operations to move other terms across the equation.
• Evaluate using a calculator for accurate results.

These steps help in efficiently finding the value of the variable, ensuring clarity and precision in your calculations.

Logarithms

Logarithms are the inverse of exponentiation, meaning that they undo the exponentiation process. They answer the question: "To what power must a base be raised, in order to achieve a given number?"
In the equation \( \log_{10}(t-3) = 2.6 \), it asks to find such a number \( t-3 \) that if \( 10 \) is raised to the power 2.6, it equals \( t-3 \). Logarithms are incredibly useful in solving equations where the exponent is unknown or needs isolating.

• They simplify problems involving exponential growth or decay.
• They are used in fields ranging from science to finance.

Understanding logarithms and their properties can greatly aid in tackling more complex mathematical problems involving exponential relationships.

Exponential Equations

Exponential equations are those where variables appear as exponents. They frequently arise in real-world contexts like compound interest, population growth, and radioactive decay. The defining feature of an exponential equation is its form: it can generally be expressed as \( a^x = b \).
These equations often need special techniques for solving, such as logarithms or rewriting them in a simpler form. For instance, rewriting \( \log_{10}(t-3) = 2.6 \) to \( 10^{2.6} = t - 3 \) simplifies the equation into one that can be solved through basic algebraic manipulation.

• Appear frequently in practical applications.
• Many require transformation into linear forms for solving.
• Simplifying these equations can make solving for variables straightforward.

A solid grasp of how to deal with exponential equations can ease understanding of many complex real-world situations and mathematical problems.