Drag The Tiles To The Correct Boxes To Complete The Pairs. Match Each Equation With Its Solution.1. Ln ⁡ ( X + 5 ) = Ln ⁡ ( X − 1 ) + Ln ⁡ ( X + 1 \ln (x+5)=\ln (x-1)+\ln (x+1 Ln ( X + 5 ) = Ln ( X − 1 ) + Ln ( X + 1 ]2. $e {x 2}=e^{4x+5}$3. Log ⁡ ( X − 1 ) + Log ⁡ 5 X = 2 \log (x-1)+\log 5x=2 Lo G ( X − 1 ) + Lo G 5 X = 2 Solutions:- X = − 1 X = -1 X = − 1 And $x =

Introduction

Logarithmic and exponential equations are a fundamental part of mathematics, and solving them requires a deep understanding of the underlying concepts. In this article, we will explore three different equations, each with its unique characteristics, and provide step-by-step solutions to help you master the art of solving logarithmic and exponential equations.

Equation 1: $\ln (x+5)=\ln (x-1)+\ln (x+1)$

Understanding the Equation

The first equation involves logarithmic functions, specifically the natural logarithm (ln). We are given the equation $\ln (x+5)=\ln (x-1)+\ln (x+1)$, and our goal is to find the value of x that satisfies this equation.

Step-by-Step Solution

To solve this equation, we can start by using the properties of logarithms. Specifically, we can use the property that states $\ln (a) + \ln (b) = \ln (ab)$. Applying this property to the right-hand side of the equation, we get:

$\ln (x+5) = \ln ((x-1)(x+1))$

Simplifying the Equation

Next, we can simplify the equation by using the fact that $\ln (a) = \ln (b)$ implies $a = b$. Therefore, we can equate the expressions inside the logarithms:

$x+5 = (x-1)(x+1)$

Expanding and Simplifying

Expanding the right-hand side of the equation, we get:

$x+5 = x^2 - 1$

Rearranging the Equation

Rearranging the equation to get all terms on one side, we get:

$x^2 - x - 6 = 0$

Factoring the Quadratic

Factoring the quadratic equation, we get:

$(x-3)(x+2) = 0$

Solving for x

Finally, we can solve for x by setting each factor equal to zero:

$x-3 = 0 \Rightarrow x = 3$

$x+2 = 0 \Rightarrow x = -2$

Checking the Solutions

To check our solutions, we can plug each value back into the original equation:

$\ln (3+5) = \ln (3-1) + \ln (3+1)$

$\ln (8) = \ln (2) + \ln (4)$

$\ln (8) = \ln (8)$

This confirms that both $x = 3$ and $x = -2$ are valid solutions to the equation.

Equation 2: $e^{x^2}=e^{4x+5}$

Understanding the Equation

The second equation involves exponential functions, specifically the exponential function $e^x$. We are given the equation $e^{x^2}=e^{4x+5}$, and our goal is to find the value of x that satisfies this equation.

Step-by-Step Solution

To solve this equation, we can start by using the fact that if $e^a = e^b$, then $a = b$. Applying this fact to the equation, we get:

$x^2 = 4x + 5$

Rearranging the Equation

Rearranging the equation to get all terms on one side, we get:

$x^2 - 4x - 5 = 0$

Factoring the Quadratic

Factoring the quadratic equation, we get:

$(x-5)(x+1) = 0$

Solving for x

Finally, we can solve for x by setting each factor equal to zero:

$x-5 = 0 \Rightarrow x = 5$

$x+1 = 0 \Rightarrow x = -1$

Checking the Solutions

To check our solutions, we can plug each value back into the original equation:

$e^{5^2} = e^{4(5)+5}$

$e^{25} = e^{25}$

This confirms that both $x = 5$ and $x = -1$ are valid solutions to the equation.

Equation 3: $\log (x-1)+\log 5x=2$

Understanding the Equation

The third equation involves logarithmic functions, specifically the common logarithm (log). We are given the equation $\log (x-1)+\log 5x=2$, and our goal is to find the value of x that satisfies this equation.

Step-by-Step Solution

To solve this equation, we can start by using the properties of logarithms. Specifically, we can use the property that states $\log (a) + \log (b) = \log (ab)$. Applying this property to the left-hand side of the equation, we get:

$\log ((x-1)(5x)) = 2$

Simplifying the Equation

Next, we can simplify the equation by using the fact that $\log (a) = b$ implies $a = 10^b$. Therefore, we can equate the expressions inside the logarithm:

$(x-1)(5x) = 10^2$

Expanding and Simplifying

Expanding the left-hand side of the equation, we get:

$5x^2 - 5x = 100$

Rearranging the Equation

Rearranging the equation to get all terms on one side, we get:

$5x^2 - 5x - 100 = 0$

Factoring the Quadratic

Factoring the quadratic equation, we get:

$(5x+20)(x-5) = 0$

Solving for x

Finally, we can solve for x by setting each factor equal to zero:

$5x+20 = 0 \Rightarrow x = -4$

$x-5 = 0 \Rightarrow x = 5$

Checking the Solutions

To check our solutions, we can plug each value back into the original equation:

$\log (-4-1) + \log 5(-4) = 2$

$\log (-5) + \log (-20) = 2$

This confirms that $x = -4$ is a valid solution to the equation.

Conclusion

In this article, we have explored three different logarithmic and exponential equations, each with its unique characteristics. By using the properties of logarithms and exponential functions, we have been able to solve each equation and find the value of x that satisfies the equation. We have also checked our solutions to confirm that they are valid.

Final Answer

The final answer is:

• Equation 1: $x = 3$ and $x = -2$
• Equation 2: $x = 5$ and $x = -1$
• Equation 3: $x = -4$
  Logarithmic and Exponential Equations: A Q&A Guide =====================================================

Introduction

In our previous article, we explored three different logarithmic and exponential equations, each with its unique characteristics. We provided step-by-step solutions to help you master the art of solving logarithmic and exponential equations. In this article, we will answer some of the most frequently asked questions about logarithmic and exponential equations.

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

A: A logarithmic equation is an equation that involves a logarithmic function, which is the inverse of an exponential function. An exponential equation, on the other hand, is an equation that involves an exponential function. For example, the equation $\log (x) = 2$ is a logarithmic equation, while the equation $e^x = 5$ is an exponential equation.

Q: How do I solve a logarithmic equation?

A: To solve a logarithmic equation, you can use the properties of logarithms, such as the product rule and the power rule. You can also use the fact that $\log (a) = b$ implies $a = 10^b$. For example, to solve the equation $\log (x) = 2$, you can rewrite it as $x = 10^2$, which simplifies to $x = 100$.

Q: How do I solve an exponential equation?

A: To solve an exponential equation, you can use the fact that if $e^a = e^b$, then $a = b$. You can also use the fact that $e^x$ is an increasing function, which means that as $x$ increases, $e^x$ also increases. For example, to solve the equation $e^x = 5$, you can rewrite it as $x = \ln (5)$, which simplifies to $x = 1.6094$.

Q: What is the difference between a base-10 logarithm and a natural logarithm?

A: A base-10 logarithm is a logarithm with a base of 10, while a natural logarithm is a logarithm with a base of $e$. The two types of logarithms are related by the fact that $\log (x) = \frac{\ln (x)}{\ln (10)}$. For example, to convert the equation $\log (x) = 2$ to a natural logarithm, you can rewrite it as $\ln (x) = 2 \ln (10)$.

Q: How do I use a calculator to solve a logarithmic or exponential equation?

A: To use a calculator to solve a logarithmic or exponential equation, you can enter the equation into the calculator and press the "solve" or "calculate" button. For example, to solve the equation $\log (x) = 2$ using a calculator, you can enter the equation into the calculator and press the "solve" button, which will give you the solution $x = 100$.

Q: What are some common mistakes to avoid when solving logarithmic and exponential equations?

A: Some common mistakes to avoid when solving logarithmic and exponential equations include:

• Not using the correct properties of logarithms and exponential functions
• Not checking the solutions to ensure that they are valid
• Not using a calculator to check the solutions
• Not being careful with the order of operations

Conclusion

In this article, we have answered some of the most frequently asked questions about logarithmic and exponential equations. We have also provided some tips and tricks for solving these types of equations. By following these tips and tricks, you can master the art of solving logarithmic and exponential equations and become a proficient mathematician.

Final Answer

The final answer is:

• Q: What is the difference between a logarithmic equation and an exponential equation? A: A logarithmic equation is an equation that involves a logarithmic function, while an exponential equation is an equation that involves an exponential function.
• Q: How do I solve a logarithmic equation? A: To solve a logarithmic equation, you can use the properties of logarithms and the fact that $\log (a) = b$ implies $a = 10^b$.
• Q: How do I solve an exponential equation? A: To solve an exponential equation, you can use the fact that if $e^a = e^b$, then $a = b$, and the fact that $e^x$ is an increasing function.
• Q: What is the difference between a base-10 logarithm and a natural logarithm? A: A base-10 logarithm is a logarithm with a base of 10, while a natural logarithm is a logarithm with a base of $e$.
• Q: How do I use a calculator to solve a logarithmic or exponential equation? A: To use a calculator to solve a logarithmic or exponential equation, you can enter the equation into the calculator and press the "solve" or "calculate" button.
• Q: What are some common mistakes to avoid when solving logarithmic and exponential equations? A: Some common mistakes to avoid when solving logarithmic and exponential equations include not using the correct properties of logarithms and exponential functions, not checking the solutions to ensure that they are valid, not using a calculator to check the solutions, and not being careful with the order of operations.