Problem 36 - Precalculus with Limits

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places. \(2^{x+1}=e^{1-x}\)

Step-by-Step Solution

Answer

The solution for x is \(x = -0.528\) when approximated to three decimal places

Step 1 - Convert \(2^{x+1}\) to exponential base e

We know that \(2 = e^{ln(2)}\), so we can rewrite \(2^{x+1}\) as \(e^{ln(2)*(x+1)}\). The initial equation would therefore be \(e^{ln(2)*(x+1)} = e^{1-x}\)

Step 2 - Equating The Exponents

Since both sides of the equation are raised to the power e, equate the exponents, which gives \(ln(2)*(x+1) = 1-x\)

Step 3 - Simplifying The Resulting Equation

Distribute the \(ln(2)\) inside the parenthesis to give \(x*ln(2) + ln(2) = 1 - x\). Then, rearrange the equation to have \(x\) terms on one side and constants on another. This gives \(x*ln(2) + x = 1 - ln(2)\)

Step 4 - Factor x...

Combine like terms on the left hand side by factoring out x, which results in \(x(ln(2) + 1) = 1 - ln(2)\)

Step 5 - Solve for x

Divide both sides of the equation by \((ln(2) + 1)\), giving \(x = \frac{1 - ln(2)}{ln(2) + 1}\)

Step 6 - Approximating to three decimal places

The numerical approximation of \(x\) to three decimal places is \(x = -0.528\)

Key Concepts

Solving Exponential Equations Algebraically

Understanding how to solve exponential equations algebraically is fundamental in algebra. An exponential equation is one in which variables appear in the exponent, like \(2^{x+1}=e^{1-x}\). The key to solving these equations lies in manipulating the equation to have the same base, which makes it simpler to solve.

In our example, the base change from 2 to \(e\), the base of natural logs, is essential. Here, using the identity \(2 = e^{\ln(2)}\), we can convert \(2^{x+1}\) to a comparable exponential form with base \(e\). This allows us to exploit the property that if \(e^{a} = e^{b}\), then \(a=b\).

After equating the exponents, we isolate \(x\) by algebraically manipulating the equation. This typically involves distributing, factoring, and ultimately isolating the variable on one side of the equation. Solving this requires comfort with algebraic operations and understanding exponential properties.

Natural Logarithm Properties

The natural logarithm, denoted as \(\ln\), is a logarithm to the base \(e\), where \(e\) is approximately equal to 2.71828. It has unique properties that make it particularly useful in solving exponential equations.

Some of these properties include:

• The \(\ln\) of a product: \(\ln(ab) = \ln(a) + \ln(b)\)
• The \(\ln\) of a quotient: \(\ln(\frac{a}{b}) = \ln(a) - \ln(b)\)
• The \(\ln\) of \(e\) to a power: \(\ln(e^{x}) = x\), which is particularly useful in our context as it allows us to equate the exponent of \(e\) to the other side.

Understanding these properties helps in transforming and simplifying exponential equations to a form where the variable can be isolated and solved for. This is evident in our solution process when we equated \(\ln(2)*(x+1)\) to \(1-x\) after converting the base.

Equating Exponents Method

The equating exponents method is a cornerstone technique when solving equations where both sides share the same base in exponential form. This strategy is based on the rule that if \(a^{m} = a^{n}\), then \(m = n\), provided that \(a\) is not equal to zero or one.

In our exercise, after converting both sides of the equation \(2^{x+1}=e^{1-x}\) to have the base \(e\), the exponents themselves can be set equal to each other because of this rule. This allows us to equate \(\ln(2)*(x+1)\) to \(1-x\) directly, bypassing the need for further exponential manipulations.

Once the exponents are equated, the algebraic operations to isolate \(x\) become straightforward; one must only follow serval algebraic steps to group, factor, and solve for the variable. As seen in the solution, by factoring out \(x\), we arrived at a solvable linear equation, which then leads to the final approximation of \(x\) to three decimal places.