Problem 58 - Intermediate Algebra

Question

Solve each equation. $$ \log 11-\log x=2 $$

Step-by-Step Solution

Answer

\( x = \frac{11}{100} \)

Step 1: Understand the Logarithmic Equation

We are given the equation \( \log 11 - \log x = 2 \). This is a logarithmic equation involving a subtraction of two logarithms.

Step 2: Apply the Logarithm Property

Use the property of logarithms that states \( \log a - \log b = \log \left( \frac{a}{b} \right) \). Apply this property to transform the equation: \[ \log \left( \frac{11}{x} \right) = 2 \].

Step 3: Exponentiate to Solve

To solve the equation \( \log \left( \frac{11}{x} \right) = 2 \), convert the logarithmic equation to an exponential equation. This means that \( \frac{11}{x} = 10^2 \).

Step 4: Simplify the Exponential Equation

Calculate \( 10^2 \): \( 100 \). Thus, the equation is \( \frac{11}{x} = 100 \).

Step 5: Solve for \( x \)

To isolate \( x \), multiply both sides of the equation by \( x \) and then divide by \( 100 \): \( 11 = 100x \). Hence, \( x = \frac{11}{100} \).

Key Concepts

Properties of Logarithms

Logarithms are powerful tools in mathematics, often used to simplify expressions of multiplication and division into addition and subtraction. One critical property is the subtraction of logarithms: \( \log a - \log b = \log \left( \frac{a}{b} \right) \). This property is a direct consequence of the definition of logarithms, which connect multiplication inside the argument with addition outside.

When we see an equation like \( \log 11 - \log x = 2 \), we can utilize this property to rewrite the entire expression as a single logarithm: \( \log \left( \frac{11}{x} \right) = 2 \). Transforming multiple terms into one can make complex equations more manageable. Understanding these properties provides insight into how logarithms work, making it easier to manipulate equations involving logarithmic expressions.

• Subtraction of Logs: Useful for combining two logarithmic terms.
• Equality of Logarithms: If \( \log_b(M) = \log_b(N) \), then \( M = N \).

Mastering these properties will deepen your problem-solving skills with logarithms, allowing you to easily navigate logarithmic equations.

Exponential Equations

Exponential equations feature variables in the exponent. Solving them often involves transforming the problem into a more accessible form. For example, with a logarithmic expression like \( \log \left( \frac{11}{x} \right) = 2 \), we exponentiate to convert it into an exponential equation.

The process of exponentiating involves recognizing that if \( \log_b(N) = c \), then \( N = b^c \). In our case, this means turning the equation into \( \frac{11}{x} = 10^2 \). By doing this, the complexity of logarithms is traded for the straightforward nature of exponents.
Some key points about exponential equations:

• Using Inverses: Transitioning between logarithmic and exponential forms uses their inverse relationship.
• Solving Power: Once in exponential form, such equations often become simple algebraic equations.

Understanding the shift from logarithms to exponentials is crucial. It allows you to switch perspectives and simplify or solve equations that might initially seem challenging.

Solving Equations

Solving equations is about finding the value of the unknown variable, which makes an equation true. In our example, once the logarithmic equation is transformed into an exponential format, we simplify and solve for \( x \).

Beginning with \( \frac{11}{x} = 100 \), the goal is to isolate \( x \). By multiplying both sides by \( x \), we eliminate the fraction, yielding \( 11 = 100x \). This algebraic form is much more straightforward to solve. We divide both sides by 100 to find \( x = \frac{11}{100} \).

• Isolation of the Variable: Crucial in solving for unknowns; isolate \( x \) to find its value.
• Verification: Plugging back into the original equation confirms the solution.

Effective equation-solving requires good understanding of algebra and the properties of exponents and logarithms. Practice and familiarity with these steps ensure adeptness at handling increasingly difficult problems.