Problem 2 - Algebra 2 and Trigonometry

Question

Pritha said that before an equation such as \(\log x=1+\log 5\) can be solved, 1 could be written as \(\log 10 .\) Do you agree with Pritha? Explain why or why not.

Step-by-Step Solution

Answer

Yes, Pritha's claim is valid; 1 can be written as \( \log 10 \).

Step 1: Understanding the Problem Statement

We are given the equation \( \log x = 1 + \log 5 \). Pritha claims that the number 1 can be written as \( \log 10 \). We need to verify if Pritha's claim is correct and how it helps in solving the equation.

Step 2: Converting 1 to a Logarithmic Form

Recall that any number \( a \) can be expressed as a logarithm of base 10 by writing it as \( \log 10^a \). Therefore, 1 can indeed be written as \( \log 10 \) because raising 10 to the power of 1 gives 10. Thus, \( 1 = \log 10 \) is a valid logarithmic expression.

Step 3: Rewriting the Equation

Using Pritha's claim, rewrite the given equation by substituting 1 with \( \log 10 \). The equation becomes \( \log x = \log 10 + \log 5 \).

Step 4: Applying Logarithmic Properties

Using the property of logarithms, \( \log a + \log b = \log (ab) \), combine the terms on the right side of the equation: \( \log 10 + \log 5 = \log (10 \times 5) = \log 50 \).

Step 5: Simplifying the Equation

Now the equation is simplified to \( \log x = \log 50 \). By the property of logarithmic equality, if \( \log a = \log b \), then \( a = b \). Thus, we conclude that \( x = 50 \).

Key Concepts

Properties of Logarithms

The properties of logarithms are fundamental tools in mathematics that simplify complex expressions and solve equations effectively. One such property is the **sum property**, which states that the sum of two logarithms with the same base can be expressed as the logarithm of the product of their arguments. In mathematical terms, \( \log a + \log b = \log (a \cdot b) \).
This property is exceptionally useful when dealing with logarithmic equations as it allows multiple logarithmic terms to be combined into one, simplifying the expressions significantly.
Another important property is the **logarithm of a power**, where \( \log (a^b) \) can be rewritten as \( b \cdot \log a \). This means a power inside a logarithm can be brought out as a multiplier.
Understanding these properties helps in converting and solving complicated logarithmic problems with ease. They not only assist in combining terms but also provide alternative paths to reach simple, solvable forms of equations.

Converting Numbers to Logarithms

Numbers can often be represented as logarithmic expressions, which is crucial when working within logarithmic equations. The basis of this conversion lies in the fact that any number can be seen as a power of another number. For instance, if you know that any number \( a \) can be expressed as \( \log 10^a \), you're using the fact that logarithms capture the concept of exponentiation.
This is evident in the given exercise, where Pritha identified that 1 can be written as \( \log 10 \) because \( 10^1 = 10 \). Hence, \( 1 = \log 10 \).
Such conversions are powerful in logarithmic equations, allowing substitution of numbers with their logarithmic equivalents to align the entire equation in a form that is more manageable. This equivalence enables the direct application of logarithmic properties to manipulate and solve equations.

Solving Logarithmic Equations

At the heart of solving logarithmic equations is simplification through the use of logarithmic properties and number conversions. The goal is to transform the equation to a point where the solution can be effortlessly deduced.
Taking the given equation \( \log x = 1 + \log 5 \), first, rewrite the number 1 as a logarithm: \( \log 10 \). This aligns the components to logarithmic terms, setting the stage for the application of properties.
The subsequent step employs the sum property, converting \( \log x = \log 10 + \log 5 \) to \( \log x = \log (10 \times 5) = \log 50 \).
Now, by leveraging the property of equality for logarithms, which states that if \( \log a = \log b \), then \( a = b \), you directly infer \( x = 50 \).
This logical sequence illustrates how strategic conversion and property application can distill a complex logarithmic problem into a straightforward solution, empowering your approach to tackling these mathematical challenges.