Question

Solve for \(x\). Give any approximate results to three significant digits. Check your answers. $$2 \log x-\log (1-x)=1$$

Short answer

After solving the equation, we get two solutions for x, where one will be extraneous and the other valid. Check each solution for validity.

Step-by-step solution

Apply the logarithm product rule

Combine the logarithms on the left side using the product rule for logarithms, which states that \(\log a - \log b = \log \frac{a}{b}\). Thus the equation becomes \(\log \frac{2x}{1-x} = 1\).

Convert to exponential form

Convert the logarithmic equation to its equivalent exponential form. Recall that if \(\log_b a = c\), then \(b^c = a\). This gives us \(10^1 = \frac{2x}{1-x}\), because \(\log\) is assumed to be base 10.

Solve for x

Now solve for \(x\). First multiply both sides by \(1-x\) to get rid of the fraction. This yields \(10(1-x) = 2x\). Next, distribute and bring all terms with \(x\) on one side and constants on the other to find the value of \(x\).

Check the result

Check the result by substituting \(x\) back into the original equation. You have to ensure that it does not lead to a logarithm of a non-positive number, because logarithms are only defined for positive arguments.

Key concepts

Product Rule for Logarithms

Understanding the product rule for logarithms is essential when trying to simplify complex logarithmic expressions. In mathematical terms, this rule can be expressed as \( \log_b (MN) = \log_b M + \log_b N \), where \(b\) is the base of the logarithm and \(M\) and \(N\) are positive real numbers.

This property suggests that the logarithm of a product equals the sum of the logarithms of the individual factors. However, in our exercise, the difference of two logarithms appears, which is an extension of the product rule: \( \log_b M - \log_b N = \log_b \frac{M}{N} \). This subtraction indicates the logarithm of a quotient. By applying this rule, we simplified the given equation to a single logarithm, making it more manageable to solve.

Converting Logarithmic to Exponential Form

Converting between logarithmic and exponential forms is a vital skill to solve logarithmic equations. It's like translating between two languages but within mathematics. The conversion is based on the definition of logarithms: if you have \( \log_b a = c \), it’s equivalent to saying \( b^c = a \).

In our exercise, we started with a logarithmic equation \( \log \frac{2x}{1-x} = 1 \), where the base (\(b\)) is assumed to be 10, as is standard when the base is not written. We transformed it into exponential form to get a clearer view of the value of \(x\) in normal arithmetic, which gives us \( 10^1 = \frac{2x}{1-x} \). This conversion is a gateway to extracting the variable from the clutches of the logarithm and solving for it.

Solving Logarithmic Equations

When we're faced with a logarithmic equation, our final goal is to isolate the variable and find its value. Solving logarithmic equations, like the one provided, involves a series of strategic moves. After employing the product rule and converting to exponential form, the next step is to manipulate the equation algebraically to isolate the variable.

In this case, after converting, we remove the fraction by multiplying both sides by \(1-x\), leading to a linear equation. From there, the strategy involves typical algebraic steps: distribute, combine like terms, and isolate \(x\) on one side. Solving logarithmic equations often requires a keen eye for logarithmic properties and a firm grasp of algebraic principles.

Logarithmic Properties

Logarithms have specific properties that dictate how they behave under different operations—the product and quotient rules are just the tip of the iceberg. For instance, there is the power rule, \( \log_b (M^k) = k \cdot \log_b M \), which expresses how a logarithm interacts with exponents. Another one is the change of base formula, useful when needing to evaluate logarithms on a calculator.

It's important to note the domain of logarithmic functions: they are only defined for positive arguments. This fact played a crucial role in our solution as we had to check that our value for \(x\) did not lead to taking the logarithm of a zero or negative number. These properties are not just academic; they are practical tools for solving and understanding logarithmic equations and their applications.