91Ó°ÊÓ

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.

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 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.