Question A1. Solve The Equation: $\[\log_a(x-2) = \log_a(8-2x) - \log_a(x-4)\\]2. For Each Of The Following, Find The Value Of \[$x\$\]: A) \[$2^x + 4^x + 8^x = 584\$\] B) \[$(\sqrt{75} - \sqrt{27} - \sqrt{3})^x

Introduction

In this article, we will be solving two types of equations: logarithmic equations and exponential equations. Logarithmic equations involve logarithmic functions, while exponential equations involve exponential functions. We will start by solving a logarithmic equation and then move on to solving an exponential equation.

Solving the Logarithmic Equation

The given logarithmic equation is:

$\log_a(x-2) = \log_a(8-2x) - \log_a(x-4)$

To solve this equation, we can start by using the properties of logarithms. We know that $\log_a(x) - \log_a(y) = \log_a(\frac{x}{y})$. We can use this property to simplify the right-hand side of the equation.

$\log_a(x-2) = \log_a(\frac{8-2x}{x-4})$

Now, we can equate the expressions inside the logarithms.

$x-2 = \frac{8-2x}{x-4}$

To solve for $x$, we can start by cross-multiplying.

$(x-2)(x-4) = 8-2x$

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

$x^2 - 6x + 8 = 8-2x$

Simplifying the equation, we get:

$x^2 - 4x = 0$

Factoring out $x$, we get:

$x(x-4) = 0$

This gives us two possible solutions: $x=0$ and $x=4$. However, we need to check if these solutions are valid.

For $x=0$, we get:

$\log_a(0-2) = \log_a(8-2(0)) - \log_a(0-4)$

$\log_a(-2) = \log_a(8) - \log_a(-4)$

This is not a valid solution, since the logarithm of a negative number is not defined.

For $x=4$, we get:

$\log_a(4-2) = \log_a(8-2(4)) - \log_a(4-4)$

$\log_a(2) = \log_a(0) - \log_a(0)$

This is also not a valid solution, since the logarithm of 0 is not defined.

Therefore, we have no valid solutions for this equation.

Solving the Exponential Equation

The given exponential equation is:

$2^x + 4^x + 8^x = 584$

To solve this equation, we can start by noticing that $4^x = (2^2)^x = 2^{2x}$ and $8^x = (2^3)^x = 2^{3x}$. We can substitute these expressions into the equation.

$2^x + 2^{2x} + 2^{3x} = 584$

We can factor out $2^x$ from the left-hand side of the equation.

$2^x(1 + 2^x + 2^{2x}) = 584$

Now, we can divide both sides of the equation by $2^x$.

$1 + 2^x + 2^{2x} = \frac{584}{2^x}$

We can simplify the right-hand side of the equation by noticing that $\frac{584}{2^x} = 2^x \cdot \frac{584}{2^{2x}} = 2^x \cdot \frac{29}{2^x}$.

$1 + 2^x + 2^{2x} = 29$

Subtracting 1 from both sides of the equation, we get:

$2^x + 2^{2x} = 28$

We can factor out $2^x$ from the left-hand side of the equation.

$2^x(1 + 2^x) = 28$

Now, we can divide both sides of the equation by $2^x$.

$1 + 2^x = \frac{28}{2^x}$

We can simplify the right-hand side of the equation by noticing that $\frac{28}{2^x} = 2^x \cdot \frac{28}{2^{2x}} = 2^x \cdot \frac{7}{2^x}$.

$1 + 2^x = 7$

Subtracting 1 from both sides of the equation, we get:

$2^x = 6$

Taking the logarithm base 2 of both sides of the equation, we get:

$x = \log_2(6)$

Therefore, the value of $x$ is $\log_2(6)$.

Solving the Exponential Equation with Square Roots

The given exponential equation is:

$(\sqrt{75} - \sqrt{27} - \sqrt{3})^x = 1$

To solve this equation, we can start by simplifying the expression inside the parentheses.

$\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$

$\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}$

$\sqrt{3}$ remains the same.

Substituting these expressions into the equation, we get:

$(5\sqrt{3} - 3\sqrt{3} - \sqrt{3})^x = 1$

Simplifying the expression inside the parentheses, we get:

$(3\sqrt{3})^x = 1$

Taking the logarithm base $3\sqrt{3}$ of both sides of the equation, we get:

$x = \log_{3\sqrt{3}}(1)$

Since $3\sqrt{3}$ is a constant, we can simplify the expression inside the logarithm.

$x = \log_{3\sqrt{3}}(1) = 0$

Therefore, the value of $x$ is 0.

Conclusion

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

A: A logarithmic equation involves logarithmic functions, while an exponential equation involves exponential functions. Logarithmic equations are used to solve for the exponent of a number, while exponential equations are used to solve for the base of a number.

Q: How do I solve a logarithmic equation?

A: To solve a logarithmic equation, you can start by using the properties of logarithms. You can use the property $\log_a(x) - \log_a(y) = \log_a(\frac{x}{y})$ to simplify the equation. You can also use the property $\log_a(x) = \frac{\log_b(x)}{\log_b(a)}$ to change the base of the logarithm.

Q: How do I solve an exponential equation?

A: To solve an exponential equation, you can start by isolating the exponential term. You can use the property $a^x = b$ to rewrite the equation as $x = \log_a(b)$. You can also use the property $a^x = b$ to rewrite the equation as $x = \frac{\log_b(b)}{\log_b(a)}$.

Q: What is the difference between a base and an exponent?

A: The base of an exponential function is the number that is being raised to a power, while the exponent is the power to which the base is being raised. For example, in the equation $2^x$, the base is 2 and the exponent is $x$.

Q: How do I simplify an exponential expression?

A: To simplify an exponential expression, you can start by using the properties of exponents. You can use the property $a^m \cdot a^n = a^{m+n}$ to combine like terms. You can also use the property $(a^m)^n = a^{m \cdot n}$ to simplify the expression.

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

A: A logarithmic function is the inverse of an exponential function. While an exponential function raises a number to a power, a logarithmic function finds the power to which a number must be raised to produce a given value.

Q: How do I graph a logarithmic function?

A: To graph a logarithmic function, you can start by plotting a few points on the graph. You can use the property $\log_a(x) = \frac{\log_b(x)}{\log_b(a)}$ to change the base of the logarithm. You can also use the property $\log_a(x) = \frac{\ln(x)}{\ln(a)}$ to graph the function.

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. While a natural logarithm is used in many mathematical and scientific applications, a common logarithm is used in many engineering and technical applications.

Q: How do I solve an exponential equation with square roots?

A: To solve an exponential equation with square roots, you can start by simplifying the expression inside the parentheses. You can use the property $\sqrt{a} = b$ to rewrite the equation as $a = b^2$. You can also use the property $\sqrt{a} = b$ to rewrite the equation as $a = b \cdot \sqrt{b}$.

Q: What is the difference between a rational exponent and an irrational exponent?

A: A rational exponent is an exponent that can be expressed as a fraction, while an irrational exponent is an exponent that cannot be expressed as a fraction. While a rational exponent can be simplified using the properties of exponents, an irrational exponent cannot be simplified using the properties of exponents.

Q: How do I simplify an exponential expression with rational exponents?

A: To simplify an exponential expression with rational exponents, you can start by using the properties of exponents. You can use the property $a^m \cdot a^n = a^{m+n}$ to combine like terms. You can also use the property $(a^m)^n = a^{m \cdot n}$ to simplify the expression.

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

A: A logarithmic equation is an equation that involves logarithmic functions, while an exponential equation with rational exponents is an equation that involves exponential functions with rational exponents. While a logarithmic equation is used to solve for the exponent of a number, an exponential equation with rational exponents is used to solve for the base of a number.

Q: How do I solve an exponential equation with rational exponents?

A: To solve an exponential equation with rational exponents, you can start by isolating the exponential term. You can use the property $a^x = b$ to rewrite the equation as $x = \log_a(b)$. You can also use the property $a^x = b$ to rewrite the equation as $x = \frac{\log_b(b)}{\log_b(a)}$.