Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Solve for \(x .\) Hint: \(\log _{a} b=c \Leftrightarrow a^{c}=b\). $$ \log _{5}(x+3)-\log _{5} x=1 $$

Step-by-Step Solution

Answer

The solution for \(x\) is \(\frac{3}{4}\).

Step 1: Apply Logarithmic Identity

Start by recognizing that the expression \( \log_5(x+3) - \log_5 x \) can be simplified using the logarithmic identity for subtraction. The formula is \( \log_a b - \log_a c = \log_a \left(\frac{b}{c}\right) \). Therefore, we can rewrite the left side of the equation as: \[\log_5 \left( \frac{x+3}{x} \right) = 1\]

Step 2: Convert Logarithm to Exponential Form

Using the hint provided, convert the logarithmic equation \( \log_5 \left( \frac{x+3}{x} \right) = 1 \) into exponential form. The equivalent expression is:\[5^1 = \frac{x+3}{x}\]Which simplifies to:\[5 = \frac{x+3}{x}\]

Step 3: Solve for x

To solve \( \frac{x+3}{x} = 5 \), start by cross-multiplying to clear the fraction. This results in:\[5x = x+3\]Next, isolate \(x\) on one side by subtracting \(x\) from both sides:\[5x - x = 3\]Simplify:\[4x = 3\]Finally, solve for \(x\) by dividing both sides by 4:\[x = \frac{3}{4}\]

Key Concepts

Logarithmic Identities

Logarithmic identities are useful tools for simplifying complex logarithmic expressions. They can help you solve equations more easily by reducing them to simpler forms. One of the most common identities is the subtraction identity. It states that the difference between two logs with the same base can be expressed as a single log of a quotient:\[ \log_a b - \log_a c = \log_a \left( \frac{b}{c} \right) \]In our problem, this identity helps us to simplify \( \log_5(x+3) - \log_5 x \) into \( \log_5 \left( \frac{x+3}{x} \right) \). This reduction is crucial because it changes a complex equation into something more manageable. By breaking it down, we keep everything within the same logarithmic framework, making it easier to apply further mathematical operations.

Exponential Form

When dealing with logarithmic equations, converting them into exponential form is a vital step. It allows you to handle equations without logarithms, simplifying the process of solving them. The basic principle here is that if you have \( \log_a b = c \), then you can convert it into its exponential form as \( a^c = b \).The problem gives us \( \log_5 \left( \frac{x+3}{x} \right) = 1 \). By applying the exponential form conversion, this becomes:\[ 5^1 = \frac{x+3}{x} \]Breaking down mathematics this way helps visualize what the equation represents in a non-logarithmic sense, which is usually simpler to solve. Exponential forms are often more intuitive for calculations and estimations, making them an essential step in problem-solving.

Solving Equations

Solving equations, especially those involving logarithms, involves applying various algebraic rules to isolate the variable. Once we've simplified the logarithmic equation into an exponential form, solving for \(x\) becomes a straightforward algebraic task.For \( \frac{x+3}{x} = 5 \), cross-multiplying helps to clear the fraction:\[ 5x = x + 3 \]This is an important step because it moves us from a complex equation involving fractions to a linear equation, which is easier to process. Next, we isolate \(x\) by subtracting \(x\) from both sides:\[ 5x - x = 3 \]Eventually, it simplifies to:\[ 4x = 3 \]Solving for \(x\) involves dividing both sides by 4:\[ x = \frac{3}{4} \]Through each step, simplifying and isolating the variable is the ultimate goal. This methodical approach to solving helps maintain accuracy and clarity, ensuring we find the correct solution to the problem.