Problem 23 - Precalculus : Building Concepts and Connections

Question

Solve the exponential equation. Round to three decimal places, when needed. $$5 e^{x}+7=32$$

Step-by-Step Solution

Answer

The solution to the exponential equation \(5 e^{x}+7=32\) is approximately x = 1.609 when rounded to three decimal places.

Step 1: Isolate the exponential term

First, isolate the exponential term on one side of the equation. Now, subtract 7 from both sides of the equation, which simplifies the equation to: \(5 e^{x} = 32 - 7 = 25\)

Step 2: Divide both side by 5

Next, divide both sides of the equation by 5 to further isolate \(e^x\). This simplifies the equation to: \(e^{x} = 25/5 = 5\)

Step 3: Apply natural logarithm on both sides

Then apply the natural logarithm, which is the inverse of the exponential function, on both sides of the equation to find the numerical value of x. The equation becomes: \(x = log(5)\)

Step 4: Round to three decimal places (if necessary)

Computing the logarithm of 5 and round to three decimal places (if the value is not an integer). Calculations result in x = 1.609

Key Concepts

Natural Logarithm

When solving exponential equations, the natural logarithm is a powerful tool. It helps us undo the exponential operation. Simply put, the natural logarithm (often denoted as \(\ln(x)\)) is the inversion of raising a base of \(e\) to a power. The number \(e\) itself is a mathematical constant, approximately 2.718, which arises in the study of growth processes and compounding interest.

If you have an equation like \(e^x = a\), you apply the natural logarithm to both sides to get \(x = \ln(a)\). This is because \(\ln(e^x) = x\), by the property of logarithms.

This process makes it possible to "undo" the \(e\) and solve for \(x\). In our exercise, for example, we've applied \(\ln\) to \(e^x = 5\) to find out that \(x = \ln(5)\). This is a key step in solving exponential equations, making the natural logarithm an indispensable tool in mathematics.

Isolating Variables

In any algebraic equation, one of the crucial steps is isolating the variable you are solving for, typically represented as \(x\). Isolating means rearranging the equation so that \(x\) stands alone on one side of the equation. This is done by reversing operations that are affecting \(x\).

Consider the original equation \(5 e^{x} + 7 = 32\). To isolate \(e^x\), start by removing constants added or subtracted from \(x\) by doing the inverse operation. Here, you subtract 7 from both sides, leading to \(5 e^{x} = 25\).

Next, remove coefficients or multipliers using division. Divide both sides by 5 to finally isolate \(e^x\), resulting in \(e^{x} = 5\).

This systematic process of isolating variables simplifies the equation, making it easier to solve for \(x\). This approach is universally applicable across different algebraic scenarios.

Rounding Decimal Places

Rounding decimal places is often needed in mathematics to simplify a number to a desired precision. It aids in making numbers more manageable and easier to read, especially in practical applications or when a precise value is not crucial.

In our exercise, once we determine that \(x = \ln(5)\), we calculate its exact value which is approximately 1.6094379. For the sake of simplicity and adhering to instructions, we round it to three decimal places.

To round to three decimal places, observe the fourth decimal place. If it is 5 or higher, round the third decimal place up by one. In our case, \(x\) becomes 1.609, since the fourth decimal place is 4 which doesn't require rounding up.

This precision of three decimal places ensures that while our value remains close to the true value, it is succinct and easy to use in further calculations.