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Today, we will learn how to solve logarithmic equations by converting them into exponential form. Can anyone remind me how we convert from logarithm to exponential?
Is it by using the formula where if log base a of b equals c, then it's a^c equals b?
Exactly! Now, let’s look at our first example: `log2(x) = 5`. How do we convert this?
We rewrite it as `x = 2^5`.
Correct! And what is `2^5`?
That’s 32! So `x = 32`.
Well done! Remember, rewriting logarithmic equations this way allows us to easily find the variable.
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Now, let's move to our second example: `log3(x + 1) = 2`. How would you solve this?
We change it to exponential form: `x + 1 = 3^2`. That gives us `x + 1 = 9`.
Great work! Now, what do we do next?
We subtract 1 to find `x = 8`.
Correct! This shows how important it is to convert and simplify effectively. Now, let’s tackle another example.
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Finally, let’s solve `log(x) + log(x - 3) = 1`. How can we simplify this using properties of logarithms?
We can apply the product rule, which means `log[x(x - 3)] = 1`.
Exactly! What does this imply in exponential form?
It means `x(x - 3) = 10`.
Awesome! Now, how do we solve this quadratic equation?
We expand it to `x^2 - 3x - 10 = 0` and factor it.
Right! Can you give me the roots?
We get `x = 5 or x = -2`. But `-2` is not valid because we cannot have negative arguments in a logarithm.
Great explanation! You've showcased how the properties of logarithms help solve complex equations effectively. Well done, everyone!
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In this section, we explore how to solve logarithmic equations by converting them into exponential form or applying logarithmic properties. Three detailed examples are presented to illustrate the steps and reasoning behind the solutions.
In this section, we delve into solving logarithmic equations, highlighting how to efficiently approach these problems using concepts covered earlier in the chapter, particularly the relationship between logarithms and exponents. A series of examples showcases the process of converting logarithmic equations into exponential form, which is a crucial step for finding the unknown variable.
Key Points:
1. Example 1: To solve the equation log2(x) = 5
, we convert to exponential form as follows:
x = 2^5
This results in x = 32
.
log3(x + 1) = 2
is transformed into exponential form:x + 1 = 3^2
, leading to x = 9 - 1
, and thus x = 8
.
log(x) + log(x - 3) = 1
, we utilize the product rule of logarithms:log[x(x - 3)] = 1
Resulting in x(x - 3) = 10
, ultimately solved as a quadratic equation, leading to values x = 5
or x = -2
(with the latter being discarded as logarithms cannot be negative).
Through these examples, we reinforce the importance of understanding logarithmic properties and the relationship with exponentials, laying the groundwork for more advanced applications of logarithmic functions in problem-solving.
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🔹 Example 3:
Solve: log𝑥 + log(𝑥 − 3) = 1
In this problem, we have two logarithmic expressions combined: log𝑥 and log(𝑥 - 3). They are set equal to 1. To solve this equation, we will use the Laws of Logarithms, specifically the Product Rule, which states that the sum of two logarithms is equal to the logarithm of the product of their arguments.
Think of it like combining two ingredients in a recipe. If you know how much of each ingredient you need when kept separately (log𝑥 for ingredient X and log(𝑥 - 3) for ingredient Y), you can combine them into one final product equation (the total amount needed).
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Use product rule:
log[𝑥(𝑥−3)] = 1
By applying the Product Rule, we can rewrite the equation log𝑥 + log(𝑥 - 3) as log[𝑥(𝑥 - 3)] = 1. This transformation makes it easier to solve for x since we have a single logarithmic expression rather than two separate ones.
Imagine you are mixing two paints together to create one color. Instead of keeping them separate (log𝑥 and log(𝑥 - 3)), you now have one mixed color (log[𝑥(𝑥 - 3)]). This makes it easier to work with and find the final outcome.
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→ 𝑥(𝑥−3) = 10^1 = 10
After we have log[𝑥(𝑥 - 3)] = 1, we convert this logarithmic equation into exponential form. The equivalent exponential form indicates that the product 𝑥(𝑥 - 3) is equal to 10 (since 10 raised to the power of 1 is 10). This step transitions us from logarithmic language back to exponential language, allowing us to solve for x more directly.
If log is like asking the question 'how much of something do I have?', converting to exponential form is like answering that question. For example, if you have a combined paint worth of 10 units, you can directly look for the amounts of the two types of paint (x and x - 3) that give you that total.
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→ 𝑥^2 - 3𝑥 - 10 = 0
We are now tasked with forming a quadratic equation. From 𝑥(𝑥 - 3) = 10, we expand it to create the equation 𝑥^2 - 3𝑥 - 10 = 0. Creating this quadratic equation allows us to use the quadratic formula or factoring to find the values of x that satisfy this equation.
Consider that you are setting up a scenario in an archery game to hit a specific target position (the number 10) with two arrows (x and x - 3). The combination of your arrows’ total ranges, represented by the quadratic, will help you evaluate the valid positions to hit that target.
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→ (𝑥−5)(𝑥+2) = 0
→ 𝑥 = 5 or 𝑥 = −2
Factoring the quadratic gives us two possible solutions: 𝑥 = 5 or 𝑥 = −2. This means that our equation has two critical points, both of which we need to analyze to check if they are valid solutions in the context of the original logarithmic expressions.
Imagine you are looking for two different paths to reach a target in a game. Both paths (their lengths represented by potential x values) seem valid initially, but you need to determine if both can lead you successfully towards your ultimate goal (the logarithmic context).
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But 𝑥 = −2 is not valid (log of a negative number is undefined),
Final answer: 𝑥 = 5
Since logarithms are undefined for negative numbers, we discard the solution x = -2. The only valid solution we keep is x = 5. This important step reveals how mathematical definitions constrain possible solutions, ensuring we only consider those that make sense in the logarithmic context.
Think about your possible paths again. If one path leads to an untraversable zone (like negative experience), you’d have to abandon that option. In contrast, a path leading to the target score (x = 5) remains viable and open for you.
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
Logarithmic Equation: An equation involving logarithms that can be solved for the variable.
Exponential Conversion: The process of rewriting a logarithmic equation in exponential form to facilitate solving.
Quadratic Solutions: Some logarithmic equations can lead to quadratic equations requiring factoring or the quadratic formula.
Valid Domains: When solving logarithmic equations, it’s crucial to ensure all solutions are valid within the logarithmic function’s domain.
See how the concepts apply in real-world scenarios to understand their practical implications.
Example 1: For log2(x) = 5
, convert to exponential form → x = 32
.
Example 2: For log3(x + 1) = 2
, convert to exponential form → x = 8
.
Example 3: For log(x) + log(x - 3) = 1
, use the product rule to solve, resulting in x = 5
.
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
When logs combine with a product fine, you'll sum them up and see them shine.
A wise old mathematician named Loggy always said that with logs, you can solve any equation, just follow the path of change and blend together what you see!
PEMA: Product, Exponential, Mixed (solution). Use this to recall the steps in solving logarithmic equations!
Review key concepts with flashcards.
Review the Definitions for terms.
Term: Logarithm
Definition:
The exponent to which a base must be raised to produce a given number.
Term: Exponential Form
Definition:
The representation of a logarithmic equation in the format a^b = c.
Term: Product Rule
Definition:
A logarithmic property stating that log(ab) = log(a) + log(b).
Term: Quadratic Equation
Definition:
An equation of the form ax² + bx + c = 0, where a, b, and c are constants.