Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Solve for \(x .\) Hint: \(\log _{a} b=c \Leftrightarrow a^{c}=b\). $$ \log _{2}(x+3)-\log _{2} x=2 $$

Step-by-Step Solution

Answer

The solution is \( x = 1 \).

Step 1: Apply the Logarithm Property

Recall that the property of logarithms states: \[ \log_b(m) - \log_b(n) = \log_b\left(\frac{m}{n}\right) \]Apply this property to the given equation:\[ \log_2(x+3) - \log_2 x = \log_2\left(\frac{x+3}{x}\right) \]Now the equation simplifies to:\[ \log_2\left(\frac{x+3}{x}\right) = 2 \]

Step 2: Convert Logarithm to Exponential Form

Using the hint provided, convert the logarithmic equation to an exponential form. The hint is: \[ \log_a b = c \Leftrightarrow a^c = b \]For our equation:\[ \log_2\left(\frac{x+3}{x}\right) = 2 \rightarrow 2^2 = \frac{x+3}{x} \]Thus, we have:\[ 4 = \frac{x+3}{x} \]

Step 3: Solve the Equation for x

Start with the equation from the previous step:\[ 4 = \frac{x+3}{x} \]Multiply both sides by \( x \) to eliminate the fraction:\[ 4x = x + 3 \]Subtract \( x \) from both sides to get:\[ 4x - x = 3 \]This simplifies to:\[ 3x = 3 \]Finally, divide both sides by 3 to solve for \( x \):\[ x = 1 \].

Key Concepts

Logarithm Properties

Logarithms have properties that make computations simpler. They help transform complex expressions into manageable parts. One essential property is the "quotient rule" for logarithms.

The quotient rule states that if you subtract one logarithm from another, with the same base, you can combine them into one logarithm. Specifically, \( \log_b(m) - \log_b(n) = \log_b\left(\frac{m}{n}\right) \).

• This means you can take two numbers, divide them, and apply a single logarithm to the result.
• It simplifies equations and helps you focus on solving the main parts of the problem rather than getting lost in combinations of terms.

In the original problem, we used this property to transform the equation \( \log_2(x+3) - \log_2 x \) into \( \log_2\left(\frac{x+3}{x}\right) \).
This step reduces complexity and makes it easier to switch to the exponential form.

Exponential Form

Converting logarithmic equations to exponential form often clarifies the solution pathway.The conversion follows the basic principle that \( \log_a(b) = c \) translates to \( a^c = b \).

• This form helps visualize and manipulate the relationship between numbers in exponential terms.
• It's particularly useful in solving equations because it "gets rid of" the logarithm, leaving a straightforward mathematical relationship.

In the example, \( \log_2\left(\frac{x+3}{x}\right) = 2 \), is converted to \( 2^2 = \frac{x+3}{x} \).

This translates the problem so that you can easily work with the equation without the logarithmic terms. It gives a clear path forward for finding the unknown value, \( x \).

Solving Equations

After converting the logarithmic to exponential form, you solve the equation.Here, your task is to unravel the expression that you derived earlier.

So, you start with \( 4 = \frac{x+3}{x} \). By multiplying both sides by \( x \), you eliminate the fraction:\( 4x = x + 3 \).

• This step is crucial as it clears the equation of fractions, simplifying further operations.
• Continue simplifying by gathering like terms, eventually leading to a quick solve for \( x \).

By subtracting \( x \) from both sides, you then have:\( 3x = 3 \).
Upon dividing both sides by 3, solving for \( x \) becomes straightforward and gives us \( x = 1 \).This step-by-step reduction is key to understanding how to navigate and solve equations effectively.