Problem 15 - Algebra 2 and Trigonometry

Question

Find \(x\) to the nearest hundredth. \(\log x-2=\log 5\)

Step-by-Step Solution

Answer

\(x = 500.00\)

Step 1: Understand the Equation

The given equation is \( \log x - 2 = \log 5 \). Our goal is to find the value of \( x \) that satisfies this equation.

Step 2: Isolate the Logarithm

Add 2 to both sides of the equation to isolate the logarithm on the left side. This gives us \( \log x = \log 5 + 2 \).

Step 3: Simplify the Equation

Recall that adding a number outside a logarithm can be rewritten as a multiplication inside the logarithm. The equation \( \log x = \log 5 + 2 \) can be rewritten, using the property \( a + b \log c = \log(c^b) \), as \( \log x = \log(5 \times 10^2) \).

Step 4: Exponentiate to Remove the Logarithm

Since the bases of the logarithms are equal, we can equate the arguments: \( x = 5 \times 100 \). Simplify the right side to get \( x = 500 \).

Step 5: Final Answer

We need to round \( x \) to the nearest hundredth. Since \( x = 500 \) is already a whole number, the rounded value is 500.00.

Key Concepts

Logarithm Properties

Logarithms are mathematical tools that help us solve equations where the variable is in an exponent. They have several important properties that make manipulating and solving these equations possible. One fundamental property is that of combining or breaking apart logarithms:

• \( \log(a \times b) = \log a + \log b \)
• \( \log\left(\frac{a}{b}\right) = \log a - \log b \)
• \( a \cdot \log b = \log(b^a) \)

In the given equation, \( \log x - 2 = \log 5 \), we look at how the "-2" outside the logarithm can be interpreted as the logarithm of a fraction. This is useful because it lets us remove the linear term and simplify the expression within the logarithmic structure.

Exponentiation

Exponentiation is the process of raising a number to a power. In the context of logarithms, it is the inverse operation. After using logarithm properties to simplify an expression, exponentiation helps us solve the equation by getting rid of the logarithm. If we have an equation like \( \log x = \log y \), by the property of equality of logarithms, we know that \( x = y \). Thus, exponentiation allows us to reframe the logarithmic equation into a simpler algebraic one. In our exercise, once we have simplified the expression to \( \log x = \log(5 \times 10^2) \), exponentiation helps us equate the objects inside the logarithm without further logs, leading us to \( x = 500 \). This step is crucial in transitioning from a logarithmic to an arithmetic form, enabling straightforward computations.

Isolation of Variables

Solving equations often involves isolating the variable we are solving for. When dealing with logarithmic equations, this means getting the variable all by itself on one side of the equation. In our original problem, isolation is used to simplify \( \log x - 2 = \log 5 \) by moving all non-logarithmic components to the opposite side. Here's how it works:

• Add 2 to both sides: \( \log x = \log 5 + 2 \)
• Express \( 2 \) as \( \log 10^2 \) using the properties of logarithms (a numeric term can be seen as a logarithm with base 10), hence simplifying to \( \log x = \log 500 \)

By the property of logarithms, once isolated, the inside of the logarithm can be directly equated, leading us through exponentiation to the solution \( x = 500 \). Isolation of variables frames the problem into a recognizable format, allowing for logical application of mathematical rules.