Problem 12 - Calculus

Question

Solve for \(x\) using logs. $$2^{x}=e^{x+1}$$

Step-by-Step Solution

Answer

x = -\frac{\ln(2)}{\ln(2) - 1}.

Step 1: Take natural log of both sides

\(\ln(2^x) = \ln(e^{x+1})\)
\(x\ln 2 = x + 1\)

Step 2: Solve for x

\(x\ln 2 - x = 1\)
\(x(\ln 2 - 1) = 1\)
\(x = \frac{1}{\ln 2 - 1}\)
Since \(\ln 2 \approx 0.693\), \(x = \frac{1}{0.693 - 1} = \frac{1}{-0.307} \approx -3.259\)

Key Concepts

Natural logarithm

The natural logarithm, denoted as \( \ln \), is a special type of logarithm that uses the base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. This type of logarithm is particularly useful in mathematics because of its unique properties and its natural occurrence in many growth processes and calculations involving exponential functions.

The property \( \ln(e) = 1 \) is very significant because it greatly simplifies expressions where \( e \) is involved. For any expression involving \( e^x \), the natural logarithm helps us easily find the exponent because:

• It "undoes" the exponential function.
• \( \ln(e^{x}) = x \) due to the inverse nature of logarithms concerning their respective exponential bases.

This is why in the original exercise, we can easily use \( \ln \) to handle the \( e^{x+1} \) term. The identity \( \ln(e^b) = b \cdot \ln(e) \) simplifies the right-hand side significantly to just \( x+1 \). By understanding the natural logarithm, solving expressions that involve the natural base \( e \) becomes simpler.

Solving exponential equations

Solving an exponential equation involves finding the unknown exponent within the equation. In the example \( 2^x = e^{x+1} \), both sides of the equation are exponential expressions with different bases – 2 and \( e \).

The strategy is to convert these exponentials into a form where their exponents can be compared or combined in a useful way. By taking the natural logarithm on both sides, we effectively strip the variables from their exponential envelopes, exposing a linear equation that is much easier to handle.

• Start by applying \( \ln \) to both sides of the equation.
• Using the property \( \ln(a^b) = b \cdot \ln(a) \), express it in terms of their exponents and logarithms of their bases.

For our example, applying \( \ln \) gives us \( \ln(2^x) = \ln(e^{x+1}) \). This simplifies to \( x \cdot \ln(2) = (x+1) \). Then, you solve this resulting linear equation for \( x \). This shows the power of logarithms in transforming seemingly complex exponential problems into simpler linear problems.

Properties of logarithms

Logarithms have several helpful properties that make them extremely powerful in solving mathematical problems involving exponential equations. Here are some of the key properties applied during the solution process:

• **Product Property:** \( \log_b(MN) = \log_b(M) + \log_b(N) \)
• **Quotient Property:** \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \)
• **Power Property:** \( \log_b(M^p) = p \cdot \log_b(M) \) as seen in the simplification \( \ln(2^x) = x \cdot \ln(2) \).
• **Change of Base Formula:** This allows the conversion from one base to another, though not directly applied here, it underlies flexibility in solving different forms of equations.

In the context of the exercise, the Power Property is crucial. It allows the exponent to "come down" and makes the expression a simple algebraic equation, easing the path to solve for \( x \). Understanding these properties enhances our ability to break down logarithmic expressions logically and systematically.