| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | B. \frac{1}{49} | B |
| deepseekr1 | To solve for \( b \) in the equation \( \log_{b}343 = -\frac{3}{2} \), we first convert the logarithmic form to its exponential equivalent: \[ b^{-\frac{3}{2}} = 343. \] Next, express \( 343 \) as \( 7^3 \): \[ b^{-\frac{3}{2}} = 7^3. \] Raise both sides to the power of \(-\frac{2}{3}\) to isolate \( b \): \[ \left(b^{-\frac{3}{2}}\right)^{-\frac{2}{3}} = \left(7^3\right)^{-\frac{2}{3}}. \] Simplifying the exponents on both sides: \[ b^{\left(-\frac{3}{2} \times -\frac{2}{3}\right)} = 7^{3 \times -\frac{2}{3}}, \] \[ b^1 = 7^{-2}. \] This simplifies to: \[ b = \frac{1}{7^2} = \frac{1}{49}. \] Verification confirms that \( \log_{\frac{1}{49}}343 = -\frac{3}{2} \) holds true. Thus, the correct answer is **B**. **Answer: B** |
B |
| gpto1 | B | B |
| GPT-4o | To solve for \( b \) if \( \log_{b}343=-\frac{3}{2} \), we can rewrite the logarithmic equation in its exponential form. The equation \( \log_{b}343 = -\frac{3}{2} \) translates to: \[ b^{-\frac{3}{2}} = 343 \] Next, to isolate \( b \), we can rewrite the equation: \[ b^{-\frac{3}{2}} = 343 \] Taking the reciprocal of both sides to get rid of the negative exponent: \[ b^{\frac{3}{2}} = \frac{1}{343} \] Now, express \( 343 \) as a power of 7: \[ 343 = 7^3 \] Thus, \[ b^{\frac{3}{2}} = \frac{1}{7^3} = 7^{-3} \] Equating the exponents, we get: \[ b^{\frac{3}{2}} = 7^{-3} \] Therefore, taking the reciprocal of both sides again to find \( b \): \[ b = 7^{-2} \] Simplifying: \[ b = \frac{1}{7^2} = \frac{1}{49} \] Thus, the correct answer is: B. \(\frac{1}{49}\) |
B |
| llama3.1-70B-Instruct | B |
B |
| llama3.1-405B-Instruct | C |
B |
| DeepSeek-Coder-V2 | C | B |
| Qwen2-72B-Instruct | B. \frac{1}{49} | B |
| baichuan4 | D | B |
| doubao | We can rewrite 343 as 7^3. So the equation becomes $\log_{b}7^3=-\frac{3}{2}$. Then $3\log_{b}7=-\frac{3}{2}$. $\log_{b}7=-\frac{1}{2}$. Rewrite it in exponential form: $b^{-\frac{1}{2}}=7$. So $\frac{1}{\sqrt{b}}=7$, $b=\frac{1}{49}$. So the answer is B. |
B |
| minimax | To find $b$ given that $\log_{b}343=-\frac{3}{2}$, we can rewrite the logarithmic equation in exponential form. The equation $\log_{b}343=-\frac{3}{2}$ means that $b$ raised to the power of $-\frac{3}{2}$ equals 343. So, we have: $$ b^{-\frac{3}{2}} = 343 $$ To solve for $b$, we can take the reciprocal of both sides and then raise both sides to the power of $-\frac{2}{3}$: $$ b^{\frac{3}{2}} = \frac{1}{343} $$ $$ (b^{\frac{3}{2}})^{\frac{2}{3}} = \left(\frac{1}{343}\right)^{\frac{2}{3}} $$ $$ b = \left(\frac{1}{343}\right)^{\frac{2}{3}} $$ Since $343 = 7^3$, we can rewrite the equation as: $$ b = \left(\frac{1}{7^3}\right)^{\frac{2}{3}} $$ $$ b = \frac{1}{(7^3)^{\frac{2}{3}}} $$ $$ b = \frac{1}{7^{3 \cdot \frac{2}{3}}} $$ $$ b = \frac{1}{7^2} $$ $$ b = \frac{1}{49} $$ Therefore, the correct answer is: B. $\frac{1}{49}$ |
B |
| kimi | To solve the equation $\log_{b}343=-\frac{3}{2}$, we can convert the logarithmic equation to its exponential form: $b^{-\frac{3}{2}} = 343$ Since $343 = 7^3$, we can rewrite the equation as: $b^{-\frac{3}{2}} = 7^3$ Now, we want to find the value of $b$ that satisfies this equation. We can rewrite the exponent on the left side to have a positive exponent by taking the reciprocal of both sides: $b^{\frac{3}{2}} = \frac{1}{7^3}$ $b^{\frac{3}{2}} = \frac{1}{343}$ Now, we raise both sides to the power of $\frac{2}{3}$ to solve for $b$: $b = \left(\frac{1}{343}\right)^{\frac{2}{3}}$ $b = \left(\frac{1}{7^3}\right)^{\frac{2}{3}}$ $b = \frac{1}{7^2}$ $b = \frac{1}{49}$ So the correct answer is: B. $\frac{1}{49}$ |
B |
| wenxinyiyan | B ### Explanation: Given the equation $\log_{b}343 = -\frac{3}{2}$, we can convert it to exponential form as follows: $b^{-\frac{3}{2}} = 343$ Since $343 = 7^3$, we can write: $b^{-\frac{3}{2}} = 7^3$ Taking the reciprocal and square root on both sides, we get: $b^{\frac{3}{2}} = \frac{1}{7^3}$ $b^{\frac{3}{2}} = \left(\frac{1}{7}\right)^3$ $b^{\frac{3}{2}} = \left(\frac{1}{7}\right)^{2 \times \frac{3}{2}}$ $b = \left(\left(\frac{1}{7}\right)^2\right)^{\frac{3}{2} \times \frac{2}{3}}$ $b = \left(\frac{1}{49}\right)^1$ $b = \frac{1}{49}$ |
B |
| zhipu | To solve the equation \(\log_{b}343 = -\frac{3}{2}\), we need to rewrite it in exponential form. The logarithmic equation \(\log_{b}343 = -\frac{3}{2}\) is equivalent to: \[ b^{-\frac{3}{2}} = 343 \] We know that \(343\) can be expressed as \(7^3\), so we substitute this into the equation: \[ b^{-\frac{3}{2}} = 7^3 \] Next, we simplify the equation by taking the cube root of both sides to eliminate the exponent \(-\frac{3}{2}\): \[ \left(b^{-\frac{3}{2}}\right)^{\frac{1}{3}} = \left(7^3\right)^{\frac{1}{3}} \] This simplifies to: \[ b^{-\frac{1}{2}} = 7 \] To isolate \(b\), we square both sides of the equation: \[ \left(b^{-\frac{1}{2}}\right)^2 = 7^2 \] This gives us: \[ b^{-1} = 49 \] Finally, we take the reciprocal of both sides to solve for \(b\): \[ b = \frac{1}{49} \] Thus, the correct answer is: B. \(\frac{1}{49}\) |
B |
| gpto1 | B | B |
| deepseekv3 | To solve for \( b \) in the equation \(\log_{b}343 = -\frac{3}{2}\), follow these steps: 1. **Convert the logarithmic equation to its exponential form:** \[ \log_{b}343 = -\frac{3}{2} \implies b^{-\frac{3}{2}} = 343 \] 2. **Express 343 as a power of 7:** \[ 343 = 7^3 \] 3. **Substitute back into the equation:** \[ b^{-\frac{3}{2}} = 7^3 \] 4. **Take the reciprocal to eliminate the negative exponent:** \[ b^{\frac{3}{2}} = 7^{-3} \] 5. **Express both sides with the same exponent:** \[ b^{\frac{3}{2}} = \left(7^{-2}\right)^{\frac{3}{2}} \] 6. **Since the exponents are equal, set the bases equal:** \[ b = 7^{-2} = \frac{1}{49} \] **Answer:** \(\boxed{B}\) |
B |