What Is The Solution Of $3 E^{-2x} + 5 = 23$?A. { -0.9$}$B. { -1.1$}$C. { -0.5$}$D. ${ 0.9\$}

Introduction

In this article, we will explore the solution of a given equation involving an exponential function. The equation is $3 e^{-2x} + 5 = 23$, and we will use algebraic manipulations to isolate the variable $x$. This type of equation is commonly encountered in mathematics, particularly in calculus and differential equations.

Understanding the Equation

The given equation involves an exponential function with a base of $e$ and an exponent of $-2x$. The constant term is $5$, and the right-hand side is $23$. Our goal is to solve for the variable $x$.

Isolating the Exponential Term

To isolate the exponential term, we first subtract $5$ from both sides of the equation:

$$3 e^{-2x} = 23 - 5$$

$$3 e^{-2x} = 18$$

Dividing by 3

Next, we divide both sides of the equation by $3$ to isolate the exponential term:

$$e^{-2x} = \frac{18}{3}$$

$$e^{-2x} = 6$$

Taking the Natural Logarithm

To eliminate the exponential function, we take the natural logarithm of both sides of the equation:

$$\ln(e^{-2x}) = \ln(6)$$

Using the property of logarithms that $\ln(e^x) = x$, we can simplify the left-hand side:

$$-2x = \ln(6)$$

Solving for x

Finally, we divide both sides of the equation by $-2$ to solve for $x$:

$$x = -\frac{\ln(6)}{2}$$

Evaluating the Expression

To evaluate the expression, we can use a calculator to find the value of $\ln(6)$:

$$\ln(6) \approx 1.79$$

Substituting this value into the expression, we get:

$$x \approx -\frac{1.79}{2}$$

$$x \approx -0.895$$

Conclusion

In this article, we solved the equation $3 e^{-2x} + 5 = 23$ using algebraic manipulations. We isolated the exponential term, divided by $3$, took the natural logarithm, and finally solved for $x$. The solution is $x \approx -0.895$, which is closest to option A.

Final Answer

The final answer is: $\boxed{-0.9}$

Introduction

In the previous article, we solved the equation $3 e^{-2x} + 5 = 23$ using algebraic manipulations. In this article, we will address some frequently asked questions (FAQs) about the given equation.

Q: What is the base of the exponential function in the given equation?

A: The base of the exponential function in the given equation is $e$, which is a mathematical constant approximately equal to $2.71828$.

Q: What is the significance of the exponent $-2x$ in the given equation?

A: The exponent $-2x$ in the given equation represents the rate of change of the exponential function. The negative sign indicates that the function decreases as $x$ increases.

Q: How do I isolate the exponential term in the given equation?

A: To isolate the exponential term, you can subtract $5$ from both sides of the equation and then divide both sides by $3$.

Q: What is the natural logarithm, and how is it used in the solution?

A: The natural logarithm is the logarithm of a number to the base $e$. It is used to eliminate the exponential function in the given equation. The natural logarithm is denoted by $\ln(x)$.

Q: How do I evaluate the expression $-\frac{\ln(6)}{2}$?

A: To evaluate the expression $-\frac{\ln(6)}{2}$, you can use a calculator to find the value of $\ln(6)$ and then divide it by $2$.

Q: What is the approximate value of $x$ in the solution?

A: The approximate value of $x$ in the solution is $-0.895$.

Q: Which option is closest to the solution?

A: Option A is closest to the solution.

Q: What is the final answer to the given equation?

A: The final answer to the given equation is $\boxed{-0.9}$.

Q: Can I use other methods to solve the given equation?

A: Yes, you can use other methods to solve the given equation, such as graphing or numerical methods. However, the method used in this article is a straightforward algebraic approach.

Q: What is the significance of the given equation in real-world applications?

A: The given equation is a simple example of an exponential function and can be used to model real-world phenomena, such as population growth or chemical reactions.

Q: Can I use the given equation to solve other problems?

A: Yes, you can use the given equation as a template to solve other problems involving exponential functions.

Conclusion

In this article, we addressed some frequently asked questions (FAQs) about the given equation $3 e^{-2x} + 5 = 23$. We provided explanations and examples to help readers understand the solution and its applications.