Problem 31 - Precalculus with Limits

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places. \(7-2 e^{x}=5\)

Step-by-Step Solution

Answer

The solution to the equation is \(x = 0\).

Step 1: Subtract 7 from both sides

Subtract 7 from both sides to get \(-2 e^{x} = 5-7\)

Step 2: Simplify Right Side

Simplify 5-7 to get \(-2\), thus the equation becomes \(-2 e^{x} = -2\)

Step 3: Divide both sides by -2

Divide every term by -2, this gives \(e^{x} = 1\)

Step 4: Calculate \(x\)

Since \(e^{x} = 1\), \(x\) is equal to the natural logarithm of 1, which is 0, thus \(x = \ln 1 = 0\)

Key Concepts

Natural Logarithms

Natural logarithms are logarithms with a special base called 'e'. The number 'e' is approximately equal to 2.71828 and has unique mathematical properties that make it very useful in calculus, exponential growth calculations, and more.

When we talk about natural logarithms, we're referring to the logarithmic function \(\ln(x)\), which is the inverse of the exponential function \(e^x\). This means that if you have an equation like \(e^x = a\), you can solve for \(x\) by taking the natural logarithm: \(x = \ln(a)\).

Some important points about natural logarithms:

• The natural logarithm of 1, \(\ln(1)\), is always 0 because \(e^0 = 1\).
• Natural logarithms help to "undo" the exponentiation by 'e', giving us the exponent that 'e' must be raised to, in order to get a specific number.
• It's crucial to understand \(ln(e) = 1\) as well, since \(e^1 = e\).

Understanding these properties is key to solving exponential equations efficiently.

Solving Exponential Equations

Solving exponential equations often involves isolating the exponential term and then applying the natural logarithm. These equations feature variables in the exponent, such as \(e^x\).

Here's a basic process for solving such equations:

• First, manipulate the equation algebraically to get the exponential term by itself on one side of the equation.
• Once isolated, take the natural logarithm of both sides, which allows you to solve for the variable in the exponent.
• Utilize properties of logarithms and exponents to simplify and solve for the unknown.

In the provided solution, an exponential equation \(7 - 2e^x = 5\) is successfully solved by first isolating \(e^x\) and then using the fact that \(e^x = 1\) leads to \(x = \ln(1) = 0\). This step-by-step process ensures you systematically reduce the complexity of the problem.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions using algebraic properties and operations. This is vital for transforming a complex expression into a simpler, solvable form.

In the context of solving exponential equations, algebraic manipulation might include:

• Moving terms from one side of the equation to the other to isolate exponential expressions, as seen in \(-2e^x = 5 - 7\). Here, constants and coefficients are moved to simplify the exponential term.
• Dividing both sides by a constant, like dividing by -2, to simplify the multiplication affecting the exponential term, resulting in \(e^x = 1\).

Each of these steps requires careful attention to maintaining equality, ensuring that each algebraic operation is performed correctly on both sides of the equation. Mastery of algebraic manipulation not only aids in solving equations but also provides clarity and structure in mathematical problem-solving.