7.4. Solve

🔹 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).

🔹 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.

🔹 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!

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