Solving Logarithmic Equations A Step By Step Guide To Solving Logₓ 1000 - Log X² + (10⁻³)^⁰ = 0
This article delves into the step-by-step solution of the logarithmic equation logₓ 1000 - log x² + (10⁻³)^⁰ = 0. We will explore the fundamental logarithmic properties and algebraic manipulations required to arrive at the solution. This detailed explanation aims to provide a clear understanding of the underlying concepts and techniques involved in solving such equations.
1. Understanding Logarithmic Equations
Logarithmic equations are equations where the logarithm of an expression containing a variable appears. To solve these equations, we often use the properties of logarithms to simplify the equation and isolate the variable. The key properties we'll use include:
- The definition of a logarithm: logₐ b = c is equivalent to aᶜ = b.
- The power rule of logarithms: logₐ bⁿ = n logₐ b.
- The logarithm of 1: logₐ 1 = 0 (for any valid base a).
- The property a⁰ = 1: Any non-zero number raised to the power of 0 is 1.
Before we dive into the solution, let's consider the importance of understanding logarithmic equations. These equations are not merely abstract mathematical concepts; they have significant applications in various fields, including physics, engineering, computer science, and finance. Logarithms help us model and analyze phenomena that exhibit exponential growth or decay, such as population growth, radioactive decay, and the magnitude of earthquakes. In computer science, logarithms are used in algorithm analysis to determine the efficiency of searching and sorting algorithms. In finance, they are crucial for calculating compound interest and analyzing investment growth.
The ability to solve logarithmic equations is therefore a fundamental skill for anyone pursuing a career in these fields. It requires a solid understanding of logarithmic properties, algebraic manipulations, and problem-solving strategies. By mastering these concepts, you can unlock a powerful toolset for tackling real-world problems.
2. Breaking Down the Equation
Let's start by dissecting the given equation: logₓ 1000 - log x² + (10⁻³)^⁰ = 0. We can identify three main terms:
- logₓ 1000: This is a logarithm with base x and argument 1000. Our first goal will be to express 1000 as a power of x.
- log x²: This is a logarithm with base 10 (assuming the base is not explicitly written) and argument x². We will use the power rule of logarithms to simplify this term.
- (10⁻³)^⁰: This is a constant term. Any non-zero number raised to the power of 0 equals 1. This will simplify this term considerably.
Now, let's look at each term individually. The first term, logₓ 1000, is where we need to find the value of x such that x raised to some power equals 1000. This highlights the fundamental relationship between logarithms and exponents. To effectively deal with this, we need to think about how we can express 1000 as a power, potentially using prime factorization or recognizing common powers.
The second term, log x², introduces the concept of the power rule of logarithms. This rule allows us to bring the exponent down as a coefficient, which can greatly simplify the equation. This is a crucial step in solving logarithmic equations, as it helps to isolate the variable and make the equation more manageable.
The third term, (10⁻³)^⁰, might seem complicated at first glance, but it's a straightforward application of the rule that any non-zero number raised to the power of 0 is 1. This simplification can significantly reduce the complexity of the equation, making it easier to solve. Understanding and applying this rule is essential for simplifying expressions and solving equations efficiently.
3. Simplifying the Equation
Our first step is to simplify each term in the equation logₓ 1000 - log x² + (10⁻³)^⁰ = 0.
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Term 1: logₓ 1000
We can rewrite 1000 as 10³, so the term becomes logₓ 10³. If we assume that x = 10, this simplifies to log₁₀ 10³ = 3. However, we need to consider a more general approach to solve for x. Let's keep this in mind and proceed with further simplification.
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Term 2: log x²
Using the power rule of logarithms, we can rewrite this term as 2 log x. This is a crucial step in simplifying the equation, as it allows us to isolate the variable and potentially combine it with other logarithmic terms.
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Term 3: (10⁻³)^⁰
Any non-zero number raised to the power of 0 is 1, so this term simplifies to 1. This simplification significantly reduces the complexity of the equation and makes it easier to solve.
Now, let's substitute these simplified terms back into the original equation. The equation becomes:
logₓ 10³ - 2 log x + 1 = 0
This simplified equation is much easier to work with. We have eliminated the exponent in the third term and rewritten the second term using the power rule of logarithms. However, we still need to address the first term, logₓ 10³, which has a variable base. To proceed further, we need to express all the logarithmic terms in the same base. This will allow us to combine them and isolate the variable.
4. Changing the Base (If Necessary)
In the simplified equation logₓ 10³ - 2 log x + 1 = 0, we have two different logarithmic terms: logₓ 10³ and log x. The first term has a variable base x, while the second term has an implied base of 10. To combine these terms, we need to express them in the same base.
We can use the change of base formula to convert logₓ 10³ to base 10. The change of base formula states:
logₐ b = logₓ b / logₓ a
Applying this formula to logₓ 10³, we get:
logₓ 10³ = log₁₀ 10³ / log₁₀ x
Now, we can simplify log₁₀ 10³ using the property logₐ aⁿ = n:
log₁₀ 10³ = 3 log₁₀ 10 = 3
So, logₓ 10³ becomes 3 / log₁₀ x. Substituting this back into our equation, we get:
(3 / log₁₀ x) - 2 log x + 1 = 0
Now, all the logarithmic terms are in base 10. To make the equation more manageable, let's introduce a substitution. Let y = log₁₀ x. The equation then becomes:
(3 / y) - 2y + 1 = 0
This equation is now a rational equation in terms of y. We have successfully transformed the original logarithmic equation into a more familiar form that we can solve using algebraic techniques. The next step is to eliminate the fraction and rearrange the equation into a standard quadratic form. This will allow us to use the quadratic formula or factoring to find the values of y.
5. Solving the Equation
Now we have the equation (3 / y) - 2y + 1 = 0. To solve this, we first eliminate the fraction by multiplying the entire equation by y:
y * [(3 / y) - 2y + 1] = y * 0
This gives us:
3 - 2y² + y = 0
Rearranging the terms, we get a quadratic equation:
2y² - y - 3 = 0
To solve this quadratic equation, we can use the quadratic formula or try to factor it. Let's try factoring:
We are looking for two numbers that multiply to (2 * -3 = -6) and add up to -1. These numbers are -3 and 2. So, we can rewrite the middle term as -3y + 2y:
2y² - 3y + 2y - 3 = 0
Now, we can factor by grouping:
y(2y - 3) + 1(2y - 3) = 0
(2y - 3)(y + 1) = 0
This gives us two possible solutions for y:
- 2y - 3 = 0 => y = 3/2
- y + 1 = 0 => y = -1
We have found the values of y, but we need to find the values of x. Recall that we made the substitution y = log₁₀ x. So, we need to solve for x in each case.
For y = 3/2:
log₁₀ x = 3/2
Using the definition of logarithms, we have:
x = 10^(3/2) = 10^(1.5) = 10 * √10 ≈ 31.62
For y = -1:
log₁₀ x = -1
Using the definition of logarithms, we have:
x = 10^(-1) = 1/10 = 0.1
So, we have found two potential solutions for x: 10^(3/2) and 0.1. However, we need to check these solutions in the original equation to ensure they are valid.
6. Verifying the Solutions
We have obtained two potential solutions for x: x = 10^(3/2) and x = 0.1. It's crucial to verify these solutions in the original equation logₓ 1000 - log x² + (10⁻³)^⁰ = 0 to ensure they are valid and do not introduce any undefined terms (such as the logarithm of a negative number or zero, or a logarithm with a non-positive base).
Let's start by verifying x = 10^(3/2):
Substitute x = 10^(3/2) into the original equation:
log₁₀^(³/₂) 1000 - log (10^(³/₂))² + (10⁻³)^⁰ = 0
We know that 1000 = 10³, so the first term becomes:
log₁₀^(³/₂) 10³ = 3 / log₁₀^(³/₂) 10 = 3 / (3/2) = 2
The second term simplifies to:
log (10^(³/₂))² = log 10³ = 3
The third term, as we know, is:
(10⁻³)^⁰ = 1
So, the equation becomes:
2 - 3 + 1 = 0
0 = 0
Thus, x = 10^(3/2) is a valid solution.
Now, let's verify x = 0.1:
Substitute x = 0.1 into the original equation:
log₀.₁ 1000 - log (0.1)² + (10⁻³)^⁰ = 0
We can rewrite 0.1 as 10⁻¹, so the first term becomes:
log₁₀⁻¹ 1000 = log₁₀⁻¹ 10³ = 3 / log₁₀⁻¹ 10 = 3 / (-1) = -3
The second term simplifies to:
log (0.1)² = log (10⁻¹)² = log 10⁻² = -2
The third term remains:
(10⁻³)^⁰ = 1
So, the equation becomes:
-3 - (-2) + 1 = 0
-3 + 2 + 1 = 0
0 = 0
Therefore, x = 0.1 is also a valid solution.
7. Final Solution
After solving the logarithmic equation and verifying the solutions, we have found two valid solutions for x:
- x = 10^(3/2) ≈ 31.62
- x = 0.1
These are the values of x that satisfy the original equation logₓ 1000 - log x² + (10⁻³)^⁰ = 0. This comprehensive step-by-step solution demonstrates the application of logarithmic properties, algebraic manipulations, and the importance of verifying solutions in the original equation.
By carefully applying the properties of logarithms and using algebraic techniques, we were able to solve this equation effectively. Remember to always verify your solutions to ensure they are valid and do not introduce any undefined terms. This meticulous approach is key to success in solving logarithmic equations and other mathematical problems.
