TA
Okay, so I need to figure out how to solve this problem where I have to find the value of x in the equation log base 2 of (x + 1) equals 3. Let me start by recalling what logarithms mean. A logarithm like log₂(x + 1) = 3 is asking the question: "2 raised to what power gives us x + 1?" Because that's the definition of a logarithm. So, if I rewrite this equation in exponential form, it should be 2^3 = x + 1. Let me check that. Yeah, because log_b(a) = c is equivalent to b^c = a. So that makes sense.
Alright, so 2^3 is 8. So then the equation becomes 8 = x + 1. To solve for x, I just need to subtract 1 from both sides. So 8 - 1 is 7. Therefore, x should be 7. Let me verify that. If I plug x = 7 back into the original equation, log₂(7 + 1) = log₂(8). And since 2^3 = 8, log₂(8) is indeed 3. So that checks out. Wait, but did I consider all the properties of logarithms here? Let me think. The logarithm is only defined for positive arguments, so x + 1 has to be greater than 0. If x = 7, then x + 1 is 8, which is positive, so that's okay. There's no issue with the domain here.
Is there another way this problem could be approached? Maybe using logarithm rules? Let's see. The equation is straightforward, so converting to exponential form seems the most direct method. I don't think there's a need to use other properties like product to sum or anything like that here. It's a simple logarithmic equation.
Wait, just to make sure I didn't make any mistakes in the algebra. Starting with log₂(x + 1) = 3. Convert to exponential: 2^3 = x + 1. 2^3 is 8, so 8 = x + 1. Subtract 1: x = 7. Yeah, that seems right. And plugging back in, log₂(7 + 1) = log₂(8) = 3. Perfect.
I guess another thing to consider is whether the logarithm could have multiple solutions, but since the logarithm function is one-to-one, there should only be one solution. So x = 7 is the only answer. Let me also recall that logarithmic functions are only defined for positive real numbers, so x + 1 > 0 implies x > -1. Since 7 is greater than -1, it's within the domain.
Hmm, maybe I can graph both sides of the equation to visualize. The left side is a logarithmic curve starting at x = -1, increasing slowly, and the right side is a horizontal line at y = 3. The intersection should be at x = 7. That makes sense.
Alternatively, if I had a different base, say base 10, the process would be the same, just converting to 10^3. But here it's base 2. So the key step is remembering how to convert from logarithmic to exponential form.
I think that's all. No steps skipped or miscalculations. Answer is 7.
To solve the equation (\log_2(x + 1) = 3):
Convert the logarithmic equation to its exponential form:
[
2^3 = x + 1
]
Simplify the exponential expression:
[
8 = x + 1
]
Solve for (x) by subtracting 1 from both sides:
[
x = 8 - 1 = 7
]
Verify the solution:
Substitute (x = 7) back into the original equation:
[
\log_2(7 + 1) = \log_2(8) = 3
]
The left side equals the right side, confirming the solution is correct.
https://www.18122.com/url/usdt/usdt3243.png
Answer: (\boxed{7})
页:
[1]