问题


There is a single choice question about high school mathematics. Answer the question by replying A, B, C or D.
Question: Evaluate $\log_8 2$.
A. \frac{1}{4}
B. 4
C. 3
D. \frac{1}{3}
Answer:

答案

模型 模型答案 正确答案
金声玉亮2.0

D. \frac{1}{3}
D
deepseekr1

To evaluate \(\log_8 2\), we need to determine the exponent \(x\) such that \(8^x = 2\). Since \(8 = 2^3\), substitute this into the equation:

\[
(2^3)^x = 2 \implies 2^{3x} = 2^1.
\]

Equating the exponents (as the bases are identical) gives \(3x = 1\), so \(x = \frac{1}{3}\).

Alternatively, using the change of base formula:

\[
\log_8 2 = \frac{\log_2 2}{\log_2 8} = \frac{1}{3}.
\]

Both methods confirm the result is \(\frac{1}{3}\).

Answer: D
D
gpto1 D D
GPT-4o To evaluate \(\log_8 2\), we need to find the exponent \(x\) such that \(8^x = 2\).

We know that \(8\) can be expressed as a power of \(2\):
\[ 8 = 2^3 \]

Thus, the equation \(8^x = 2\) can be rewritten as:
\[ (2^3)^x = 2 \]

Using the properties of exponents, we get:
\[ 2^{3x} = 2^1 \]

Since the bases are the same, we can set the exponents equal to each other:
\[ 3x = 1 \]

Solving for \(x\):
\[ x = \frac{1}{3} \]

Therefore, \(\log_8 2 = \frac{1}{3}\).

Answer: D
D
llama3.1-70B-Instruct

A
D
llama3.1-405B-Instruct

D
D
DeepSeek-Coder-V2 D. \frac{1}{3} D
Qwen2-72B-Instruct D. \frac{1}{3} D
baichuan4 D D
doubao D. $\frac{1}{3}$.

We can rewrite $2$ as $8^{\frac{1}{3}}$, so $\log_8 2=\log_8 8^{\frac{1}{3}}=\frac{1}{3}$.
D
minimax To evaluate $\log_8 2$, we need to determine the exponent to which 8 must be raised to get 2.

Since $8 = 2^3$, we can rewrite the logarithm in terms of base 2:

$\log_8 2 = \log_{2^3} 2$

Using the property of logarithms that $\log_{a^b} c = \frac{1}{b} \log_a c$, we get:

$\log_{2^3} 2 = \frac{1}{3} \log_2 2$

Since $\log_2 2 = 1$ (because $2^1 = 2$), we have:

$\frac{1}{3} \log_2 2 = \frac{1}{3} \cdot 1 = \frac{1}{3}$

Therefore, the answer is:

D. $\frac{1}{3}$
D
kimi To evaluate $\log_8 2$, we can use the change of base formula for logarithms, which states that $\log_b a = \frac{\log_c a}{\log_c b}$ for any positive numbers $a$, $b$, and $c$ (where $a \neq 1$ and $b \neq 1$). Using base 10 (common logarithm), we get:

$$\log_8 2 = \frac{\log_{10} 2}{\log_{10} 8}$$

We know that $\log_{10} 8 = \log_{10} (2^3) = 3 \log_{10} 2$. So,

$$\log_8 2 = \frac{\log_{10} 2}{3 \log_{10} 2} = \frac{1}{3}$$

Therefore, the correct answer is:

D. $\frac{1}{3}$
D
wenxinyiyan To evaluate $\log_8 2$, we need to find the number $x$ such that $8^x = 2$.

Recall that $8 = 2^3$. We can rewrite the equation as $(2^3)^x = 2$.

Using the power of a power rule, we have $2^{3x} = 2$.

Since the base is the same on both sides, we can set the exponents equal: $3x = 1$.

Solving for $x$, we get $x = \frac{1}{3}$.

Therefore, $\log_8 2 = \frac{1}{3}$.

The answer is D. $\frac{1}{3}$.
D
zhipu To evaluate \(\log_8 2\), we need to determine the power to which 8 must be raised to get 2.

Recall that \(8 = 2^3\). Therefore, we can rewrite the logarithm in terms of base 2:

\[
\log_8 2 = \log_{2^3} 2
\]

Using the change of base formula, \(\log_b a = \frac{\log_c a}{\log_c b}\), we can change the base to 2:

\[
\log_{2^3} 2 = \frac{\log_2 2}{\log_2 (2^3)}
\]

We know that \(\log_2 2 = 1\) because \(2^1 = 2\), and \(\log_2 (2^3) = 3\) because \(2^3 = 8\).

Thus,

\[
\log_{2^3} 2 = \frac{1}{3}
\]

So, the correct answer is:

D. \(\frac{1}{3}\)
D
gpto1 D D
deepseekv3 To evaluate \(\log_8 2\), we need to determine the exponent \(x\) such that:

\[
8^x = 2
\]

We know that \(8\) can be written as \(2^3\). Substituting this into the equation gives:

\[
(2^3)^x = 2
\]

Simplifying the left side:

\[
2^{3x} = 2^1
\]

Since the bases are the same, we can set the exponents equal to each other:

\[
3x = 1
\]

Solving for \(x\):

\[
x = \frac{1}{3}
\]

Therefore, the value of \(\log_8 2\) is \(\frac{1}{3}\).

**Answer:** \(\boxed{D}\)
D