Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 37-40, use natural logarithms to solve each of the exponential equations. Hint: To solve \(3^{x}=11\), take ln of both sides, obtaining \(x \ln 3=\ln 11\); then \(x=(\ln 11) /(\ln 3) \approx 2.1827\). \(2^{x}=17\)

Step-by-Step Solution

Answer

\(x \approx 4.0875\)

Step 1: Understand the given equation

The problem asks us to solve the exponential equation \(2^{x} = 17\).

Step 2: Take the natural logarithm of both sides

Apply the natural logarithm (\(\ln\)) to both sides of the equation. This gives us \(\ln(2^{x}) = \ln(17)\).

Step 3: Apply the power rule of logarithms

Use the power rule of logarithms, which states that \(\ln(a^{b}) = b \cdot \ln(a)\). This transforms our equation to \(x \cdot \ln(2) = \ln(17)\).

Step 4: Solve for \(x\)

Isolate \(x\) by dividing both sides of the equation by \(\ln(2)\). This results in \(x = \frac{\ln(17)}{\ln(2)}\).

Step 5: Calculate the value of \(x\)

Compute the values of \(\ln(17)\) and \(\ln(2)\), then perform the division. This gives approximately \(x \approx \frac{2.8332}{0.6931} \approx 4.0875\).

Key Concepts

Exponential Equations

Exponential equations are mathematical expressions where variables appear as exponents. These types of equations are common in many scientific and mathematical fields. The general form of an exponential equation is \( a^{x} = b \), where "\( a \)" is the base, "\( x \)" is the exponent, and "\( b \)" is the result. In such equations, the goal is often to find the value of the exponent.
A classic example would be solving \( 2^{x} = 17 \), where we need to find "\( x \)". Exponential equations can be solved using logarithms because they help to "bring down" the exponent, making the equation easier to manipulate.
In particular, natural logarithms, which use the constant "e" as their base, offer a powerful tool in converting the product form of exponents into a sum, simplifying the solution process.

Power Rule of Logarithms

The power rule of logarithms is a key property that facilitates solving equations, especially those involving exponents. According to this rule, for any positive number "\( a \)" and real number "\( b \)", the logarithm of "\( a \)" raised to the power "\( b \)" is \( b \times \ln(a) \).
This rule can be represented mathematically as:

• \( \ln(a^{b}) = b \cdot \ln(a) \)

The power rule allows us to move the exponent in front of the logarithm, converting a multiplicative scenario into an additive one.
For instance, in the equation \( \ln(2^{x}) = \ln(17) \), we apply the power rule to get \( x \cdot \ln(2) = \ln(17) \). By moving the variable "\( x \)" outside of the logarithm, it becomes straightforward to solve for "\( x \)" through simple division.

Solving Equations Using Logarithms

Logarithms are exceptionally useful in tackling exponential equations. They convert the multiplicative nature of exponentiation into an additive one, which is generally easier to work with.
To solve an equation like \( 2^x = 17 \), the first step is to take the natural logarithm of both sides, resulting in \( \ln(2^x) = \ln(17) \).
Next, applying the power rule of logarithms, this becomes \( x \cdot \ln(2) = \ln(17) \), making it a simple task to isolate "\( x \)".

• Divide both sides by \( \ln(2) \) to get \( x = \frac{\ln(17)}{\ln(2)} \).
• Calculate the values: \( \ln(17) \approx 2.8332 \) and \( \ln(2) \approx 0.6931 \).
• Solve: \( x \approx \frac{2.8332}{0.6931} \approx 4.0875 \).

This step-by-step process highlights the power of logarithms in simplifying and solving exponential equations effectively.