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Today, we’re going to learn how to solve logarithmic equations. First, can anyone tell me how to convert a logarithmic equation into exponential form?
Is it just changing log_a(b) = c to a^c = b?
Exactly! Well done! This is fundamental to solving logarithmic equations. Now, let's try an example together: how would you solve log_2(x) = 5?
So, that means x = 2^5, which equals 32!
Correct! Great job! Just remember that the base must be positive, and we cannot use 1.
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Now, let's look at how we can use the properties of logarithms to simplify equations. Can anyone remind me of the product rule?
Log(mn) = log(m) + log(n)!
Exactly! Let’s apply this! If we have log(x) + log(x-3) = 1, what should we do first?
We can use the product rule to combine them into log[x(x-3)] = 1!
Perfect! This simplifies our equation significantly. What would we do next?
We then convert log[x(x-3)] to exponential form!
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Let’s discuss how we find our solutions and the importance of checking for extraneous solutions. Who can summarize what we need to do after solving for x?
We should plug our answer back into the original equation to ensure it’s valid!
Great point! Can anyone think of a reason why this is important?
If the result gives us a negative logarithm, that means we did something wrong, right?
Exactly! Always check your work! Now, let’s tackle some practice problems.
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In this section, students learn to solve logarithmic equations by converting them into exponential form. The section emphasizes methods such as using logarithmic properties and solving quadratic equations that arise from logarithmic expressions.
Logarithmic equations present unique challenges that require understanding the relationship between logarithms and exponents.
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🔹 Example 1:
Solve: log 𝑥 = 5
→ Convert to exponential form:
𝑥 = 25 = 32
In this example, we have the logarithmic equation log 𝑥 = 5. To solve this equation, we need to convert it into its equivalent exponential form. The logarithmic equation states that the base raised to the exponent gives us the argument (the number inside the logarithm). Here, we know that if log 𝑥 = 5, it means that the base (which is usually 10 if no base is provided) raised to the power of 5 equals 𝑥. Therefore, we can write it as 10^5 = 𝑥. Simplifying this gives us 𝑥 = 100000. However, the example shows an alternative exponential form as x = 2^5, which is simply stating the bases in powers instead. Thus, in general, 𝑥 would equal 100000 or 2^5 based on the context provided in logarithms.
Consider a situation where you are baking and you have a recipe that says, 'To make a delicious cake, you need to bake it for 5 times at a specific temperature.' We can think of the base (temperature) as the constant that when raised to the time (5) will give us the end result (the delicious cake). In logarithms, we're trying to find out what that temperature (base) must be to get our final result (the 100000 units of cake).
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🔹 Example 2:
Solve: log (𝑥+1) = 2
→ 𝑥+1 = 32 = 9
→ 𝑥 = 8
Here we start with the equation log (𝑥 + 1) = 2. Again, we will convert this logarithmic equation to exponential form. This means the equation can be rewritten as the base raised to the power of 2 is equal to the argument (𝑥 + 1). Thus, we have 10^2 = 𝑥 + 1, which simplifies to 100 = 𝑥 + 1. From here, we can isolate 𝑥 by subtracting 1 from both sides, leading us to 𝑥 = 100 - 1 = 99 - the correct addition process through understanding how logarithmic equations reveal unknowns.
Imagine you have a jar of cookies and you know that if you add 1 more cookie to your jar, the total will be equal to multiplying an original count of cookies (say ten), two times. In logarithmic terms, what you're really trying to find out is how many cookies you originally had (𝑥). The equation shows how the power of multiplication reflects back on how many items you originally had plus the extra cookie.
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🔹 Example 3:
Solve: log𝑥 + log(𝑥 −3) = 1
Use product rule:
log[𝑥(𝑥−3)] = 1
→ 𝑥(𝑥−3) = 10^1 = 10
→ 𝑥² − 3𝑥 − 10 = 0
→ Solve quadratic: (𝑥−5)(𝑥+2) = 0
→ 𝑥 = 5 or 𝑥 = −2
But 𝑥 = −2 is not valid (log of negative number is undefined),
Final answer: 𝑥 = 5
In this example, we tackle the equation log𝑥 + log(𝑥 − 3) = 1. First, we apply the product rule of logarithms, which states that the sum of logs is the log of the product. This gives us log[𝑥(𝑥 − 3)] = 1. We then convert this logarithmic equation into an exponential one: 10^1 = 𝑥(𝑥 − 3). After simplification, we move terms around to get a quadratic equation 𝑥² − 3𝑥 − 10 = 0. We then factor this as (𝑥 -5)(𝑥 + 2) = 0, giving potential solutions 𝑥 = 5 and 𝑥 = -2. Since we can't take the logarithm of a negative number, we reject 𝑥 = -2, leading us to our final answer of 5.
Imagine you are working with discounts at a store. You can think of the equation as a scenario where one log represents one type of discount (say 5% on some items), and another log represents another item discount (let's say 3%). The combined discount (the product of the two) leads you to finding the total amount spent as log total. The quadratic in the end would be like discovering rare offers when mixing and matching discounts, where sometimes they're negative (like losing money) and sometimes they're beneficial!
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
Logarithmic Conversion: Changing log_a(b) into a^c = b.
Properties of Logarithms: Using rules like the Product Rule to simplify logarithmic expressions.
Validity of Solutions: Checking all found solutions to ensure they fit the original equation.
See how the concepts apply in real-world scenarios to understand their practical implications.
Example 1: To solve log_2(x) = 5, convert to exponential: x = 2^5 = 32.
Example 2: For log(x+1) = 2, converting gives us x+1 = 10, so x = 9.
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
Logarithm here, don’t you fear, change to the power, your answer is near.
Imagine a log cabin needing to convert logs (logarithms) into firewood (exponential results) for a warm, cozy experience, making transformations easy!
PE/V: 'Product is Plus, Exponent is Variable' to remember logging rules.
Review key concepts with flashcards.
Review the Definitions for terms.
Term: Logarithm
Definition:
The power to which a number must be raised to obtain another number.
Term: Exponential form
Definition:
The representation of logarithmic equations rewritten using exponentials.
Term: Product Rule
Definition:
log(mn) = log(m) + log(n) for any positive m and n.
Term: Extraneous solution
Definition:
A solution that does not satisfy the original equation.