Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(\frac{3000}{2+e^{2 x}}=2\)

Short answer

The solution to the equation is approximately \(x = 3.176\).

Step-by-step solution

Transpose the equation to isolate exponential term

Start by writing the equation as \(3000 = 2 * (2 + e^{2 x})\). Transposing terms gives \(1498 = e^{2*x}\).

Apply logarithm rule to isolate \(x\)

From the exponential identity \(a = b^x\) implies \(log_b(a) = x\), we can transform the equation into a simpler format to isolate \(x\), \(2*x = log(1498)\) or \(x = log(1498)/2\).

Approximate the result

Finally, for the required approximation the result needs to be evaluated up to three decimal places. Using a calculator, the solution to the equation is approximately \(x = 3.176\).

Key concepts

Algebra

Algebra is an essential area of mathematics that deals with symbols and the rules for manipulating these symbols. It provides a framework to understand how to solve equations like the one given in the exercise. In this specific problem, we start by rearranging or transposing the equation to simplify it.The original equation is \(\frac{3000}{2+e^{2x}}=2\). The goal is to isolate the exponential part. By multiplying both sides of the equation by the denominator \(2+e^{2x}\), we derive a simpler equation: \(3000 = 2 * (2 + e^{2x})\).We continue simplifying by expanding and transposing terms, which means moving them from one side of the equation to the other while maintaining equality. In doing this, we arrive at \(1498 = e^{2x}\). This is a significant simplification, as it isolates the term involving the exponent ready for the next step of using logarithms.

Logarithms

The step involving logarithms takes center stage when dealing with exponential equations. Logarithms are the inverses of exponentials and are quite handy when you need to solve equations involving exponential terms. In our equation \(1498 = e^{2x}\), the goal is to solve for \(x\). Recall the relationship between exponents and logarithms: if \(a = b^x\), then \(x = \log_b(a)\). Here, \(b\) is the base of the exponent, which is \(e\) in our case, meaning we are dealing with natural logarithms.Applying the natural logarithm on both sides, we simplify it to \(2x = \log_e(1498)\) or simply \(2x = \ln(1498)\). To isolate \(x\), divide both sides by 2 to get \(x = \frac{\ln(1498)}{2}\). Logarithms transform the exponential equation into a linear one, making it much easier to handle.

Mathematical Approximation

Mathematical approximation is crucial, especially in situations where exact answers are laborious to compute or where they contain too many digits to be practical. This is why, after isolating \(x\), it's essential to approximate the result to three decimal places.Using a calculator or computational tool, we calculate \(\ln(1498)\) and subsequently divide by 2, as per our expression \(x = \frac{\ln(1498)}{2}\). The value we arrive at is \(x \approx 3.176\).This approximation provides a practical solution, precise enough for most applications. It's an example of how approximations maintain efficiency without significantly sacrificing accuracy, especially when dealing with measurements or constants in the real world where data precision could vary.