Find The Value(s) Of X X X For Each Of The Following Equations:a) E 2 X − 3 E X + 2 = 0 E^{2x} - 3e^x + 2 = 0 E 2 X − 3 E X + 2 = 0 B) Log ⁡ 4 ( X + 1 ) − Log ⁡ 4 ( X − 1 ) = 2 \log_4(x+1) - \log_4(x-1) = 2 Lo G 4 ​ ( X + 1 ) − Lo G 4 ​ ( X − 1 ) = 2

Introduction

Exponential and logarithmic equations are fundamental concepts in mathematics, and solving them requires a deep understanding of the underlying principles. In this article, we will delve into the world of exponential and logarithmic equations, exploring the techniques and strategies required to solve them. We will focus on two specific equations: $e^{2x} - 3e^x + 2 = 0$ and $\log_4(x+1) - \log_4(x-1) = 2$. Our goal is to find the value(s) of $x$ for each of these equations.

Understanding Exponential Equations

Exponential equations involve the variable $x$ in the exponent of a base number. In the equation $e^{2x} - 3e^x + 2 = 0$, the base is $e$, and the exponent is $2x$. To solve this equation, we need to isolate the variable $x$.

Substitution Method

One approach to solving exponential equations is the substitution method. We can substitute a new variable, say $y$, for the expression $e^x$. This allows us to rewrite the equation in terms of $y$ and solve for $y$. Once we have found the value of $y$, we can substitute it back into the original equation to find the value of $x$.

Solving the First Equation

Let's apply the substitution method to the first equation: $e^{2x} - 3e^x + 2 = 0$. We can substitute $y = e^x$, which gives us:

$$y^2 - 3y + 2 = 0$$

This is a quadratic equation in $y$, which we can solve using the quadratic formula:

$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, $a = 1$, $b = -3$, and $c = 2$. Plugging these values into the quadratic formula, we get:

$$y = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(2)}}{2(1)}$$

Simplifying the expression under the square root, we get:

$$y = \frac{3 \pm \sqrt{9 - 8}}{2}$$

$$y = \frac{3 \pm \sqrt{1}}{2}$$

$$y = \frac{3 \pm 1}{2}$$

This gives us two possible values for $y$:

$$y = \frac{3 + 1}{2} = 2$$

$$y = \frac{3 - 1}{2} = 1$$

Now that we have found the values of $y$, we can substitute them back into the original equation to find the values of $x$. For $y = 2$, we have:

$$e^x = 2$$

Taking the natural logarithm of both sides, we get:

$$x = \ln(2)$$

For $y = 1$, we have:

$$e^x = 1$$

Taking the natural logarithm of both sides, we get:

$$x = \ln(1) = 0$$

Therefore, the solutions to the first equation are $x = \ln(2)$ and $x = 0$.

Understanding Logarithmic Equations

Logarithmic equations involve the variable $x$ in the argument of a logarithmic function. In the equation $\log_4(x+1) - \log_4(x-1) = 2$, the base is $4$, and the argument is $x+1$ and $x-1$. To solve this equation, we need to isolate the variable $x$.

Properties of Logarithms

One approach to solving logarithmic equations is to use the properties of logarithms. Specifically, we can use the property that states $\log_a(b) - \log_a(c) = \log_a(\frac{b}{c})$. This allows us to combine the two logarithmic terms into a single logarithmic term.

Solving the Second Equation

Let's apply the properties of logarithms to the second equation: $\log_4(x+1) - \log_4(x-1) = 2$. We can use the property mentioned above to combine the two logarithmic terms:

$$\log_4(x+1) - \log_4(x-1) = \log_4\left(\frac{x+1}{x-1}\right)$$

This gives us:

$$\log_4\left(\frac{x+1}{x-1}\right) = 2$$

We can rewrite this equation in exponential form:

$$\frac{x+1}{x-1} = 4^2$$

Simplifying the right-hand side, we get:

$$\frac{x+1}{x-1} = 16$$

Cross-multiplying, we get:

$$x+1 = 16x - 16$$

Simplifying the equation, we get:

$$15x = 17$$

Dividing both sides by $15$, we get:

$$x = \frac{17}{15}$$

Therefore, the solution to the second equation is $x = \frac{17}{15}$.

Conclusion

In this article, we have explored the techniques and strategies required to solve exponential and logarithmic equations. We have focused on two specific equations: $e^{2x} - 3e^x + 2 = 0$ and $\log_4(x+1) - \log_4(x-1) = 2$. Our goal was to find the value(s) of $x$ for each of these equations. We have used the substitution method and the properties of logarithms to solve these equations. The solutions to the first equation are $x = \ln(2)$ and $x = 0$, while the solution to the second equation is $x = \frac{17}{15}$. We hope that this article has provided a comprehensive guide to solving exponential and logarithmic equations.

Introduction

Exponential and logarithmic equations are fundamental concepts in mathematics, and solving them requires a deep understanding of the underlying principles. In this article, we will address some of the most frequently asked questions about exponential and logarithmic equations.

Q: What is the difference between exponential and logarithmic equations?

A: Exponential equations involve the variable $x$ in the exponent of a base number, while logarithmic equations involve the variable $x$ in the argument of a logarithmic function.

Q: How do I solve exponential equations?

A: To solve exponential equations, you can use the substitution method, where you substitute a new variable, say $y$, for the expression $e^x$. This allows you to rewrite the equation in terms of $y$ and solve for $y$. Once you have found the value of $y$, you can substitute it back into the original equation to find the value of $x$.

Q: How do I solve logarithmic equations?

A: To solve logarithmic equations, you can use the properties of logarithms, such as the property that states $\log_a(b) - \log_a(c) = \log_a(\frac{b}{c})$. This allows you to combine the two logarithmic terms into a single logarithmic term.

Q: What is the difference between a natural logarithm and a common logarithm?

A: A natural logarithm is a logarithm with a base of $e$, while a common logarithm is a logarithm with a base of $10$.

Q: How do I convert between natural logarithms and common logarithms?

A: To convert between natural logarithms and common logarithms, you can use the following formula:

$$\ln(x) = \frac{\log(x)}{\log(e)}$$

Q: What is the difference between a logarithmic function and an exponential function?

A: A logarithmic function is a function that involves the variable $x$ in the argument of a logarithmic function, while an exponential function is a function that involves the variable $x$ in the exponent of a base number.

Q: How do I graph logarithmic and exponential functions?

A: To graph logarithmic and exponential functions, you can use a graphing calculator or a computer program. You can also use the properties of logarithms and exponentials to graph these functions.

Q: What are some common applications of exponential and logarithmic equations?

A: Exponential and logarithmic equations have many common applications in science, engineering, and finance. Some examples include:

• Modeling population growth and decay
• Calculating interest rates and investment returns
• Analyzing data and making predictions
• Solving problems in physics and engineering

Conclusion

In this article, we have addressed some of the most frequently asked questions about exponential and logarithmic equations. We hope that this article has provided a comprehensive guide to these concepts and has helped you to better understand how to solve exponential and logarithmic equations.

Additional Resources

If you are looking for additional resources to help you learn more about exponential and logarithmic equations, we recommend the following:

• Khan Academy: Exponential and Logarithmic Equations
• Mathway: Exponential and Logarithmic Equations
• Wolfram Alpha: Exponential and Logarithmic Equations

We hope that this article has been helpful in your studies. Good luck with your math homework!