Solve The Equation: Ln ⁡ X + Ln ⁡ ( X + 35 ) = Ln ⁡ 74 \ln X + \ln (x + 35) = \ln 74 Ln X + Ln ( X + 35 ) = Ln 74

Introduction

In this article, we will delve into the world of mathematics and explore a fascinating equation involving logarithms. The equation in question is $\ln x + \ln (x + 35) = \ln 74$. Our goal is to solve for the variable $x$ and understand the underlying concepts that make this equation tick. We will break down the solution into manageable steps, making it easy to follow along and grasp the underlying mathematics.

Understanding Logarithms

Before we dive into the solution, let's take a moment to understand the concept of logarithms. A logarithm is the inverse operation of exponentiation. In other words, if we have a number $y$ and we want to find the exponent $x$ such that $a^x = y$, then the logarithm of $y$ with base $a$ is defined as $\log_a y = x$. In this article, we will be working with natural logarithms, denoted by $\ln x$, which is the logarithm of $x$ with base $e$, where $e$ is a mathematical constant approximately equal to $2.71828$.

Step 1: Combine the Logarithms

The first step in solving the equation is to combine the logarithms on the left-hand side. Using the property of logarithms that states $\log_a (m) + \log_a (n) = \log_a (mn)$, we can rewrite the equation as:

$$\ln x + \ln (x + 35) = \ln 74$$

$$\ln (x(x + 35)) = \ln 74$$

Step 2: Exponentiate Both Sides

Now that we have combined the logarithms, we can exponentiate both sides of the equation to get rid of the logarithms. Since we are working with natural logarithms, we can use the fact that $e^{\ln x} = x$ to simplify the equation:

$$e^{\ln (x(x + 35))} = e^{\ln 74}$$

$$x(x + 35) = 74$$

Step 3: Expand and Simplify

The next step is to expand and simplify the equation. We can start by multiplying out the terms on the left-hand side:

$$x^2 + 35x = 74$$

Step 4: Rearrange the Equation

Now that we have expanded and simplified the equation, we can rearrange it to get a quadratic equation in the form $ax^2 + bx + c = 0$. We can do this by subtracting $74$ from both sides of the equation:

$$x^2 + 35x - 74 = 0$$

Step 5: Solve the Quadratic Equation

The final step is to solve the quadratic equation. We can use the quadratic formula to find the solutions:

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

In this case, $a = 1$, $b = 35$, and $c = -74$. Plugging these values into the formula, we get:

$$x = \frac{-35 \pm \sqrt{35^2 - 4(1)(-74)}}{2(1)}$$

$$x = \frac{-35 \pm \sqrt{1225 + 296}}{2}$$

$$x = \frac{-35 \pm \sqrt{1521}}{2}$$

$$x = \frac{-35 \pm 39}{2}$$

This gives us two possible solutions:

$$x = \frac{-35 + 39}{2} = 2$$

$$x = \frac{-35 - 39}{2} = -37$$

Conclusion

In this article, we have solved the equation $\ln x + \ln (x + 35) = \ln 74$ using a step-by-step approach. We combined the logarithms, exponentiated both sides, expanded and simplified the equation, rearranged it to get a quadratic equation, and finally solved the quadratic equation using the quadratic formula. The two possible solutions are $x = 2$ and $x = -37$. We hope that this article has provided a clear and concise explanation of the solution to this equation.

Final Thoughts

Solving equations involving logarithms can be a challenging task, but with the right approach and techniques, it can be done. In this article, we have demonstrated how to combine logarithms, exponentiate both sides, expand and simplify the equation, rearrange it to get a quadratic equation, and finally solve the quadratic equation using the quadratic formula. We hope that this article has provided a valuable resource for students and mathematicians alike.

Additional Resources

For those who want to learn more about logarithms and quadratic equations, we recommend checking out the following resources:

• Khan Academy: Logarithms
• Khan Academy: Quadratic Equations
• Wolfram MathWorld: Logarithm
• Wolfram MathWorld: Quadratic Equation

References

• "Calculus" by Michael Spivak
• "Algebra" by Michael Artin
• "Mathematics for the Nonmathematician" by Morris Kline

Introduction

In our previous article, we solved the equation $\ln x + \ln (x + 35) = \ln 74$ using a step-by-step approach. We combined the logarithms, exponentiated both sides, expanded and simplified the equation, rearranged it to get a quadratic equation, and finally solved the quadratic equation using the quadratic formula. In this article, we will answer some frequently asked questions about the solution to this equation.

Q: What is the final answer to the equation?

A: The final answer to the equation is $x = 2$ and $x = -37$.

Q: How do I combine the logarithms in the equation?

A: To combine the logarithms in the equation, we use the property of logarithms that states $\log_a (m) + \log_a (n) = \log_a (mn)$. In this case, we have $\ln x + \ln (x + 35) = \ln 74$, which can be rewritten as $\ln (x(x + 35)) = \ln 74$.

Q: Why do I need to exponentiate both sides of the equation?

A: Exponentiating both sides of the equation is necessary to get rid of the logarithms. Since we are working with natural logarithms, we can use the fact that $e^{\ln x} = x$ to simplify the equation.

Q: How do I expand and simplify the equation?

A: To expand and simplify the equation, we can start by multiplying out the terms on the left-hand side. In this case, we have $x(x + 35) = 74$, which can be rewritten as $x^2 + 35x = 74$.

Q: Why do I need to rearrange the equation to get a quadratic equation?

A: Rearranging the equation to get a quadratic equation is necessary to solve for the variable $x$. By subtracting $74$ from both sides of the equation, we get $x^2 + 35x - 74 = 0$, which is a quadratic equation in the form $ax^2 + bx + c = 0$.

Q: How do I solve the quadratic equation?

A: To solve the quadratic equation, we can use the quadratic formula, which is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. In this case, we have $a = 1$, $b = 35$, and $c = -74$, which can be plugged into the formula to get the solutions.

Q: What are the solutions to the equation?

A: The solutions to the equation are $x = 2$ and $x = -37$.

Q: Why are there two solutions to the equation?

A: There are two solutions to the equation because the quadratic equation $x^2 + 35x - 74 = 0$ has two roots. The quadratic formula gives us two possible values for $x$, which are $x = 2$ and $x = -37$.

Q: How do I check my solutions?

A: To check your solutions, you can plug them back into the original equation to see if they satisfy the equation. In this case, we can plug $x = 2$ and $x = -37$ back into the equation $\ln x + \ln (x + 35) = \ln 74$ to see if they are true.

Conclusion

In this article, we have answered some frequently asked questions about the solution to the equation $\ln x + \ln (x + 35) = \ln 74$. We hope that this article has provided a clear and concise explanation of the solution to this equation and has helped to clarify any confusion.

Additional Resources

For those who want to learn more about logarithms and quadratic equations, we recommend checking out the following resources:

• Khan Academy: Logarithms
• Khan Academy: Quadratic Equations
• Wolfram MathWorld: Logarithm
• Wolfram MathWorld: Quadratic Equation

References

• "Calculus" by Michael Spivak
• "Algebra" by Michael Artin
• "Mathematics for the Nonmathematician" by Morris Kline

Note: The references provided are for general information purposes only and are not directly related to the solution of the equation.