Problem 32 - Precalculus with Calculus Previews

Question

Solve the given logarithmic equation. $$ \log _{2}(x-3)-\log _{2}(2 x+1)=-\log _{2} 4 $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{13}{2} \).

Step 1: Apply Logarithm Properties

The equation given is \( \log_{2}(x-3) - \log_{2}(2x+1) = -\log_{2} 4 \). We can apply the property of logarithms that states \( \log_{b}(A) - \log_{b}(B) = \log_{b}(\frac{A}{B}) \). So, our equation becomes \( \log_{2}\left(\frac{x-3}{2x+1}\right) = -\log_{2} 4 \).

Step 2: Simplify with Inverse Logarithm

We want to eliminate the logarithms by expressing them in exponential form. Since \( -\log_{2} 4 = \log_{2}(4^{-1}) \), the equation becomes \( \log_{2}\left(\frac{x-3}{2x+1}\right) = \log_{2}\left(\frac{1}{4}\right) \).

Step 3: Remove the Logarithms

Since the logarithms on both sides have the same base, we can set the arguments equal to each other: \( \frac{x-3}{2x+1} = \frac{1}{4} \).

Step 4: Cross Multiply

To remove the fractions, cross multiply: \( 4(x-3) = 2x+1 \). This simplifies to \( 4x - 12 = 2x + 1 \).

Step 5: Solve for x

Rearrange the equation to find \( x \). Subtract \( 2x \) from both sides to get \( 2x - 12 = 1 \). Add 12 to both sides to obtain \( 2x = 13 \). Finally, divide by 2 to solve for \( x \): \( x = \frac{13}{2} \).

Step 6: Verify the Solution

Substitute \( x = \frac{13}{2} \) back into the original equation to verify that it holds. Check that both \( x-3 \) and \( 2x+1 \) are positive, which they are, confirming the solution is valid.

Key Concepts

Logarithm Properties

Understanding the properties of logarithms is key to solving logarithmic equations efficiently. Logarithms are mathematical functions that help us solve for exponents. They have several properties that make operations easier to handle.

One useful property is the **quotient rule**, which states that the difference of two logarithms with the same base can be expressed as a single logarithm. For any positive numbers \( A \) and \( B \), and a base \( b \), the property is given by:

• \( \log_b(A) - \log_b(B) = \log_b(\frac{A}{B}) \)

In our exercise, this helps us transform the original equation from two separate logarithms of \( \log_2(x-3) \) and \( \log_2(2x+1) \) into a single logarithm of \( \log_2\left(\frac{x-3}{2x+1}\right) \).

Understanding and using this property simplifies equations and reduces them to a form that is easier to solve.

Exponential Form

Once you've applied the logarithm properties, converting the logarithmic equation into its exponential form is the next step to solving it. This happens when you want to "undo" the logarithm and work with exponential terms instead.

In the exercise, the equation \( \log_2\left(\frac{x-3}{2x+1}\right) = \log_2\left(\frac{1}{4}\right) \) can be simplified by noticing that both sides have the same logarithmic base. This allows us to set their arguments equal, dropping the log altogether.

We know, for example, that if \( \log_{b} y = x \), then in exponential form, this means \( b^x = y \). Therefore, if both sides of an equation are \( \log_2(...) \), we can equate the insides directly:

• \( \frac{x-3}{2x+1} = \frac{1}{4} \)

Using exponential form to remove the logarithm simplifies the comparison of numerical and algebraic expressions further.

Cross Multiplication

When fractions are present in an equation, cross multiplication is a straightforward method to eliminate them.

This technique involves multiplying the numerator of one fraction by the denominator of the other fraction, and vice versa. In the provided exercise, the equation \( \frac{x-3}{2x+1} = \frac{1}{4} \) becomes a simple matter of cross multiplying to solve for \( x \).

• You multiply \( 4(x-3) = 2x + 1 \)

This results in a manageable linear equation. Simplifying and solving linear equations through cross multiplication reduces complex fractions into more basic algebraic operations. This step is crucial as it paves the way to simplify, rearrange, and solve equations more efficiently.