Solve The Following Equations. Give Non-exact Values Correct To 3 Significant Figures.(a) $\log_2(x+6) - \log_2(x-1) = 3$(b) $\log_8 X + \log_8(3x+4) = 2$(c) $\log_5(2x+5) = 1 + \log_5(x-2$\](d) $\log_3(2x+1) + \log_3(x-1)

Introduction

Logarithmic equations are a type of mathematical equation that involves logarithms. They are used to solve problems that involve exponential growth or decay. In this article, we will solve four logarithmic equations with non-exact values correct to 3 significant figures.

Equation (a): $\log_2(x+6) - \log_2(x-1) = 3$

To solve this equation, we can use the property of logarithms that states $\log_a(b) - \log_a(c) = \log_a(\frac{b}{c})$. Applying this property to the given equation, we get:

$\log_2(x+6) - \log_2(x-1) = \log_2(\frac{x+6}{x-1}) = 3$

Now, we can rewrite the equation in exponential form:

$\frac{x+6}{x-1} = 2^3$

Simplifying the equation, we get:

$\frac{x+6}{x-1} = 8$

Cross-multiplying, we get:

$x+6 = 8x - 8$

Subtracting $x$ from both sides, we get:

$6 = 7x - 8$

Adding $8$ to both sides, we get:

$14 = 7x$

Dividing both sides by $7$, we get:

$x = 2$

However, we need to check if this value satisfies the original equation. Plugging $x = 2$ into the original equation, we get:

$\log_2(2+6) - \log_2(2-1) = \log_2(8) - \log_2(1) = 3 - 0 = 3$

Since the value $x = 2$ satisfies the original equation, we have found the solution to the equation.

Equation (b): $\log_8 x + \log_8(3x+4) = 2$

To solve this equation, we can use the property of logarithms that states $\log_a(b) + \log_a(c) = \log_a(bc)$. Applying this property to the given equation, we get:

$\log_8 x + \log_8(3x+4) = \log_8(x(3x+4)) = 2$

Now, we can rewrite the equation in exponential form:

$x(3x+4) = 8^2$

Simplifying the equation, we get:

$3x^2 + 4x = 64$

Rearranging the equation, we get:

$3x^2 + 4x - 64 = 0$

This is a quadratic equation, and we can solve it using the quadratic formula:

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

In this case, $a = 3$, $b = 4$, and $c = -64$. Plugging these values into the quadratic formula, we get:

$x = \frac{-4 \pm \sqrt{4^2 - 4(3)(-64)}}{2(3)}$

Simplifying the equation, we get:

$x = \frac{-4 \pm \sqrt{16 + 768}}{6}$

$x = \frac{-4 \pm \sqrt{784}}{6}$

$x = \frac{-4 \pm 28}{6}$

Simplifying the equation, we get:

$x = \frac{-4 + 28}{6}$ or $x = \frac{-4 - 28}{6}$

$x = \frac{24}{6}$ or $x = \frac{-32}{6}$

$x = 4$ or $x = -\frac{16}{3}$

However, we need to check if these values satisfy the original equation. Plugging $x = 4$ into the original equation, we get:

$\log_8(4) + \log_8(3(4)+4) = \log_8(4) + \log_8(16) = 2$

Since the value $x = 4$ satisfies the original equation, we have found one solution to the equation.

Plugging $x = -\frac{16}{3}$ into the original equation, we get:

$\log_8(-\frac{16}{3}) + \log_8(3(-\frac{16}{3})+4) = \log_8(-\frac{16}{3}) + \log_8(-4) = 2$

However, this value does not satisfy the original equation because the logarithm of a negative number is undefined.

Equation (c): $\log_5(2x+5) = 1 + \log_5(x-2)$

To solve this equation, we can use the property of logarithms that states $\log_a(b) - \log_a(c) = \log_a(\frac{b}{c})$. Applying this property to the given equation, we get:

$\log_5(2x+5) - \log_5(x-2) = 1$

Now, we can rewrite the equation in exponential form:

$\frac{2x+5}{x-2} = 5^1$

Simplifying the equation, we get:

$\frac{2x+5}{x-2} = 5$

Cross-multiplying, we get:

$2x+5 = 5x - 10$

Subtracting $2x$ from both sides, we get:

$5 = 3x - 10$

Adding $10$ to both sides, we get:

$15 = 3x$

Dividing both sides by $3$, we get:

$x = 5$

However, we need to check if this value satisfies the original equation. Plugging $x = 5$ into the original equation, we get:

$\log_5(2(5)+5) = \log_5(15) = 1 + \log_5(5-2)$

$\log_5(15) = 1 + \log_5(3)$

Since the value $x = 5$ does not satisfy the original equation, we have not found the solution to the equation.

Equation (d): $\log_3(2x+1) + \log_3(x-1) = 0$

To solve this equation, we can use the property of logarithms that states $\log_a(b) + \log_a(c) = \log_a(bc)$. Applying this property to the given equation, we get:

$\log_3(2x+1) + \log_3(x-1) = \log_3((2x+1)(x-1)) = 0$

Now, we can rewrite the equation in exponential form:

$(2x+1)(x-1) = 3^0$

Simplifying the equation, we get:

$(2x+1)(x-1) = 1$

Expanding the equation, we get:

$2x^2 - x - 1 = 1$

Rearranging the equation, we get:

$2x^2 - x - 2 = 0$

This is a quadratic equation, and we can solve it using the quadratic formula:

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

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

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

Simplifying the equation, we get:

$x = \frac{1 \pm \sqrt{1 + 16}}{4}$

$x = \frac{1 \pm \sqrt{17}}{4}$

Simplifying the equation, we get:

$x = \frac{1 + \sqrt{17}}{4}$ or $x = \frac{1 - \sqrt{17}}{4}$

However, we need to check if these values satisfy the original equation. Plugging $x = \frac{1 + \sqrt{17}}{4}$ into the original equation, we get:

$\log_3(2(\frac{1 + \sqrt{17}}{4})+1) + \log_3(\frac{1 + \sqrt{17}}{4}-1) = \log_3(\frac{1 + \sqrt{17}}{2}+1) + \log_3(\frac{1 + \sqrt{17}}{4}-1)$

Since the value $x = \frac{1 + \sqrt{17}}{4}$ does not satisfy the original equation, we have not found the solution to the equation.

Plugging $x = \frac{1 - \sqrt{17}}{4}$ into the original equation, we get:

$\log_3(2(\frac{1 - \sqrt{17}}{4})+1) + \log_3(\frac{1 - \sqrt{17}}{4}-1) = \log_3(\frac{1 - \sqrt{17}}{2}+1) + \log_3(\frac{1 - \sqrt{17}}{4}-1)$

Since the value $x = \frac{1 - \sqrt{17}}{4}$ does not satisfy

Introduction

In the previous article, we solved four logarithmic equations with non-exact values correct to 3 significant figures. In this article, we will answer some frequently asked questions about logarithmic equations and provide additional examples to help you understand the concepts better.

Q: What is a logarithmic equation?

A: A logarithmic equation is a type of mathematical equation that involves logarithms. It is used to solve problems that involve exponential growth or decay.

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

A: A logarithmic equation is an equation that involves logarithms, while an exponential equation is an equation that involves exponents. For example, the equation $\log_2(x+6) - \log_2(x-1) = 3$ 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 can use the properties of logarithms, such as the product rule, the quotient rule, and the power rule. You can also use the fact that $\log_a(b) = c$ is equivalent to $a^c = b$.

Q: What is the product rule for logarithms?

A: The product rule for logarithms states that $\log_a(b) + \log_a(c) = \log_a(bc)$. This means that the logarithm of the product of two numbers is equal to the sum of their logarithms.

Q: What is the quotient rule for logarithms?

A: The quotient rule for logarithms states that $\log_a(b) - \log_a(c) = \log_a(\frac{b}{c})$. This means that the logarithm of the quotient of two numbers is equal to the difference of their logarithms.

Q: What is the power rule for logarithms?

A: The power rule for logarithms states that $\log_a(b^c) = c\log_a(b)$. This means that the logarithm of a number raised to a power is equal to the power times the logarithm of the number.

Q: How do I solve an equation with a logarithm base other than 10 or e?

A: To solve an equation with a logarithm base other than 10 or e, you can use the change of base formula, which states that $\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$, where $c$ is any positive number other than 1.

Q: What is the change of base formula?

A: The change of base formula is a formula that allows you to change the base of a logarithm from one base to another. It is given by the equation $\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$, where $c$ is any positive number other than 1.

Q: How do I solve an equation with a logarithm and a fraction?

A: To solve an equation with a logarithm and a fraction, you can use the properties of logarithms, such as the product rule, the quotient rule, and the power rule. You can also use the fact that $\log_a(b) = c$ is equivalent to $a^c = b$.

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

A: A logarithmic equation is an equation that involves logarithms, while a polynomial equation is an equation that involves only polynomials. For example, the equation $\log_2(x+6) - \log_2(x-1) = 3$ is a logarithmic equation, while the equation $x^2 + 3x - 4 = 0$ is a polynomial equation.

Q: How do I solve a polynomial equation?

A: To solve a polynomial equation, you can use various methods, such as factoring, the quadratic formula, or numerical methods.

Q: What is the quadratic formula?

A: The quadratic formula is a formula that allows you to solve a quadratic equation of the form $ax^2 + bx + c = 0$. It is given by the equation $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Q: How do I solve an equation with a logarithm and a square root?

A: To solve an equation with a logarithm and a square root, you can use the properties of logarithms, such as the product rule, the quotient rule, and the power rule. You can also use the fact that $\log_a(b) = c$ is equivalent to $a^c = b$.

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

A: A logarithmic equation is an equation that involves logarithms, while a trigonometric equation is an equation that involves trigonometric functions, such as sine, cosine, and tangent. For example, the equation $\log_2(x+6) - \log_2(x-1) = 3$ is a logarithmic equation, while the equation $\sin(x) = \frac{1}{2}$ is a trigonometric equation.

Q: How do I solve a trigonometric equation?

A: To solve a trigonometric equation, you can use various methods, such as the unit circle, trigonometric identities, or numerical methods.

Q: What is the unit circle?

A: The unit circle is a circle with a radius of 1 that is centered at the origin of a coordinate plane. It is used to help solve trigonometric equations.

Q: How do I solve an equation with a logarithm and a trigonometric function?

A: To solve an equation with a logarithm and a trigonometric function, you can use the properties of logarithms, such as the product rule, the quotient rule, and the power rule. You can also use the fact that $\log_a(b) = c$ is equivalent to $a^c = b$.

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

A: A logarithmic equation is an equation that involves logarithms, while a differential equation is an equation that involves derivatives. For example, the equation $\log_2(x+6) - \log_2(x-1) = 3$ is a logarithmic equation, while the equation $\frac{dy}{dx} = 2x$ is a differential equation.

Q: How do I solve a differential equation?

A: To solve a differential equation, you can use various methods, such as separation of variables, integration, or numerical methods.

Q: What is the difference between a logarithmic equation and a system of equations?

A: A logarithmic equation is an equation that involves logarithms, while a system of equations is a set of two or more equations that are solved simultaneously. For example, the equation $\log_2(x+6) - \log_2(x-1) = 3$ is a logarithmic equation, while the system of equations $\begin{cases} x + y = 2 \\ x - y = 1 \end{cases}$ is a system of equations.

Q: How do I solve a system of equations?

A: To solve a system of equations, you can use various methods, such as substitution, elimination, or numerical methods.

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

A: A logarithmic equation is an equation that involves logarithms, while a matrix equation is an equation that involves matrices. For example, the equation $\log_2(x+6) - \log_2(x-1) = 3$ is a logarithmic equation, while the equation $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 6 \end{bmatrix}$ is a matrix equation.

Q: How do I solve a matrix equation?

A: To solve a matrix equation, you can use various methods, such as Gaussian elimination, LU decomposition, or numerical methods.

Q: What is the difference between a logarithmic equation and a linear programming problem?

A: A logarithmic equation is an equation that involves logarithms, while a linear programming problem is a problem that involves maximizing or minimizing a linear function subject to certain constraints. For example, the equation $\log_2(x+6) - \log_2(x-1) = 3$ is a logarithmic equation, while the linear programming problem $\max 2x + 3y$ subject to $x + y \leq 4$ and $x \geq 0$ and $y \geq 0$ is a linear programming problem.

Q: How do I solve a linear programming problem?

A: To solve a linear programming problem, you can use various methods, such as the simplex method, the dual simplex method, or numerical methods.

Q: What is the difference between a logarithmic equation and a nonlinear programming problem?

A: A logarithmic equation is an equation that involves logarithms, while a nonlinear programming problem is a problem that involves maximizing or minimizing a nonlinear function subject to certain constraints. For example, the equation $\log_2(x+6) - \log_2(x-1) = 3$ is a logarithmic equation, while the nonlinear programming problem $\max x^2 + 2xy + y^2$ subject to $x + y \leq 4$ and $x \geq 0