Solve Each Of The Following Equations For X X X .a. Ln ⁡ ( 6 X ) = 0 \ln (6x) = 0 Ln ( 6 X ) = 0 X = X = X = □ \square □ B. Log ⁡ 4 ( 2 X ) + Log ⁡ 4 ( 4 X ) = 2 \log_4(2x) + \log_4(4x) = 2 Lo G 4 ​ ( 2 X ) + Lo G 4 ​ ( 4 X ) = 2 X = X = X = □ \square □ C. Log ⁡ 7 ( − 3 X ) + Log ⁡ 7 ( − 10 X ) = 3 \log_7(-3x) + \log_7(-10x) = 3 Lo G 7 ​ ( − 3 X ) + Lo G 7 ​ ( − 10 X ) = 3 X = X = X =

Introduction

Logarithmic and exponential functions are essential in mathematics, and solving equations involving these functions can be challenging. In this article, we will focus on solving three equations that involve logarithms and exponents. We will use various techniques to isolate the variable and find the solution.

Equation a: Solving $\ln (6x) = 0$

The first equation we will solve is $\ln (6x) = 0$. To solve this equation, we need to use the definition of the natural logarithm function, which states that $\ln x = 0$ if and only if $x = 1$. Therefore, we can rewrite the equation as:

$$\ln (6x) = \ln 1$$

Since the logarithm function is one-to-one, we can equate the arguments of the logarithms:

$$6x = 1$$

Now, we can solve for $x$ by dividing both sides of the equation by 6:

$$x = \frac{1}{6}$$

Therefore, the solution to the equation $\ln (6x) = 0$ is $x = \frac{1}{6}$.

Equation b: Solving $\log_4(2x) + \log_4(4x) = 2$

The second equation we will solve is $\log_4(2x) + \log_4(4x) = 2$. To solve this equation, we can use the product rule of logarithms, which states that $\log_b x + \log_b y = \log_b (xy)$. Applying this rule to the given equation, we get:

$$\log_4(2x \cdot 4x) = 2$$

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

$$\log_4(8x^2) = 2$$

Now, we can rewrite the equation in exponential form:

$$4^2 = 8x^2$$

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

$$16 = 8x^2$$

Dividing both sides of the equation by 8, we get:

$$2 = x^2$$

Taking the square root of both sides of the equation, we get:

$$x = \pm \sqrt{2}$$

Therefore, the solutions to the equation $\log_4(2x) + \log_4(4x) = 2$ are $x = \sqrt{2}$ and $x = -\sqrt{2}$.

Equation c: Solving $\log_7(-3x) + \log_7(-10x) = 3$

The third equation we will solve is $\log_7(-3x) + \log_7(-10x) = 3$. To solve this equation, we can use the product rule of logarithms, which states that $\log_b x + \log_b y = \log_b (xy)$. Applying this rule to the given equation, we get:

$$\log_7(-3x \cdot -10x) = 3$$

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

$$\log_7(30x^2) = 3$$

Now, we can rewrite the equation in exponential form:

$$7^3 = 30x^2$$

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

$$343 = 30x^2$$

Dividing both sides of the equation by 30, we get:

$$\frac{343}{30} = x^2$$

Taking the square root of both sides of the equation, we get:

$$x = \pm \sqrt{\frac{343}{30}}$$

Simplifying the expression under the square root, we get:

$$x = \pm \frac{\sqrt{343}}{\sqrt{30}}$$

Simplifying the square roots, we get:

$$x = \pm \frac{7\sqrt{7}}{\sqrt{30}}$$

Rationalizing the denominator, we get:

$$x = \pm \frac{7\sqrt{7}\sqrt{30}}{\sqrt{30}\sqrt{30}}$$

Simplifying the denominator, we get:

$$x = \pm \frac{7\sqrt{210}}{30}$$

Simplifying the expression under the square root, we get:

$$x = \pm \frac{7\sqrt{6 \cdot 5 \cdot 7}}{30}$$

Simplifying the expression under the square root, we get:

$$x = \pm \frac{7\sqrt{6} \cdot \sqrt{5} \cdot \sqrt{7}}{30}$$

Simplifying the expression, we get:

$$x = \pm \frac{7\sqrt{210}}{30}$$

Therefore, the solutions to the equation $\log_7(-3x) + \log_7(-10x) = 3$ are $x = \frac{7\sqrt{210}}{30}$ and $x = -\frac{7\sqrt{210}}{30}$.

Conclusion

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Introduction

In our previous article, we solved three equations that involved logarithms and exponents. We used various techniques, including the product rule of logarithms and the definition of the natural logarithm function, to isolate the variable and find the solution. In this article, we will answer some common questions that readers may have about solving equations with logarithms and exponents.

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 $\log_2 x = 3$ is a logarithmic equation, while the equation $2^x = 8$ is an exponential equation.

Q: How do I solve a logarithmic equation?

A: To solve a logarithmic equation, you need to use the definition of the logarithmic function. The definition states that $\log_b x = y$ if and only if $b^y = x$. You can use this definition to rewrite the logarithmic equation in exponential form, and then solve for the variable.

Q: What is the product rule of logarithms?

A: The product rule of logarithms states that $\log_b x + \log_b y = \log_b (xy)$. This means that you can combine two logarithmic expressions with the same base into a single logarithmic expression.

Q: How do I use the product rule of logarithms to solve an equation?

A: To use the product rule of logarithms to solve an equation, you need to apply the rule to the equation and then simplify the resulting expression. For example, if you have the equation $\log_2 x + \log_2 y = 2$, you can use the product rule to rewrite the equation as $\log_2 (xy) = 2$. You can then rewrite the equation in exponential form and solve for the variable.

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

A: A natural logarithm is a logarithm with a base of $e$, where $e$ is a mathematical constant approximately equal to 2.718. A common logarithm, on the other hand, is a logarithm with a base of 10. The natural logarithm is often denoted by the symbol $\ln$, while the common logarithm is often denoted by the symbol $\log$.

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:

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

This formula allows you to convert a natural logarithm to a common logarithm, and vice versa.

Q: What are some common mistakes to avoid when solving equations with logarithms and exponents?

A: Some common mistakes to avoid when solving equations with logarithms and exponents include:

• Forgetting to check the domain of the logarithmic function
• Forgetting to check the range of the exponential function
• Not using the product rule of logarithms correctly
• Not using the definition of the logarithmic function correctly
• Not checking for extraneous solutions

Conclusion

In this article, we answered some common questions that readers may have about solving equations with logarithms and exponents. We covered topics such as the difference between logarithmic and exponential equations, the product rule of logarithms, and common mistakes to avoid when solving equations with logarithms and exponents. We hope that this article has been helpful in clarifying some of the concepts and techniques involved in solving equations with logarithms and exponents.