Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(\frac{400}{1+e^{-x}}=350\)

Short answer

The solution to the equation is approximately \(x \approx 1.946\).

Step-by-step solution

Isolate \(e^{-x}\)

First, rearranging the equation \( \frac{400}{1+e^{-x}} = 350\) to isolate \(e^{-x}\) gives \(e^{-x} = \frac{400}{350} - 1\).

Simplify expression

Simplify the expression \( \frac{400}{350} - 1\) to get \(e^{-x}=\frac{50}{350}\) or equivalently, \(e^{-x}=\frac{1}{7}\).

Apply natural logarithm to both sides

Take the natural logarithm of both sides of the equation \(e^{-x} = \frac{1}{7}\) to get \(-x = \ln(\frac{1}{7})\). The natural logarithm and the exponential function cancel each other out, which provides a way to get \(x\) out of the exponent.

Solve for \(x\)

Finally, solve for \(x\) by multiplying both sides of the equation \(-x = \ln(\frac{1}{7})\) by -1. This gives \(x = -\ln(\frac{1}{7})\).

Approximate the result

Using a calculator, find the numerical value of \(-\ln(\frac{1}{7})\), rounded to three decimal places, this gives \(x \approx 1.946\).

Key concepts

Natural Logarithms

Natural logarithms are a fundamental concept in mathematics that help us understand growth processes and exponential functions. The natural logarithm, often represented as \(\ln(x)\), is the logarithm to the base \(e\), where \(e\) is approximately equal to 2.71828. In essence, the natural logarithm of a number \(x\) is the power to which \(e\) must be raised to obtain \(x\). This is particularly useful in solving exponential equations where the variable is in the exponent.
For example, in the equation \(e^{-x} = \frac{1}{7}\), applying the natural logarithm to both sides allows us to "bring down" the exponent. This transforms it into \(-x = \ln(\frac{1}{7})\), a much simpler form that is easier to solve for \(x\).
Natural logarithms have certain properties that make them especially handy:

• The natural log of 1 is 0, because \(e^0 = 1\).
• \(\ln(e) = 1\) because \(e^1 = e\).
• They convert multiplication into addition: \(\ln(a \cdot b) = \ln a + \ln b\).
• They convert division into subtraction: \(\ln(\frac{a}{b}) = \ln a - \ln b\).

Understanding and using natural logarithms is a crucial skill when dealing with exponential equations.

Isolating Variables

Isolating variables is an essential technique in algebra that involves manipulating an equation so that a variable stands alone on one side. This skill is fundamental when you need to solve equations where the variable can't be directly observed. For instance, in the given exponential equation \(\frac{400}{1+e^{-x}}=350\), our primary goal is to solve for \(x\).
The first step involves isolating terms which contain the variable. In this case, we want to make \(e^{-x}\) the subject by rearranging the equation. Rearranging involves a series of arithmetic steps: first, cross-multiply the equation to eliminate the fraction, and then solve for \(e^{-x}\). This results in the expression \(e^{-x} = \frac{50}{350}\).
Simplifying further, we find \(e^{-x} = \frac{1}{7}\). Such simplifications are necessary to easily work with the variable and apply further mathematical methods like natural logarithms to eradicate the exponent. By mastering the art of isolating variables, you make solving complex equations much more tractable.

Approximation Methods

Approximation methods are used in mathematics when a precise solution is either unattainable or unnecessary, particularly with irrational numbers or complex calculations. In order to solve an equation and find a solution that is close enough for practical purposes, we often round off to a certain number of decimal places. This is what is meant by "approximation."
After finding \(x = -\ln(\frac{1}{7})\), the next step is to use a calculator to compute this value. The natural logarithm \(\ln(\frac{1}{7})\) yields an irrational number. Although precise, it will have more decimals than necessary. Rounding off to three decimal places provides a usable and clear result for most practical purposes.
In many scenarios, mathematicians and scientists choose approximations when:

• The values have no exact solution.
• The exact value is not practical for the purpose of the task.
• The task involves modeling or theoretical work, where approximations can still yield meaningful insights.

Therefore, approximating \(x\) to 1.946 in our exercise simplifies the solution without losing significant accuracy.