Problem Analysis #15

K-EXAONE-236B-A23B · Run 1

Correct

70.11s

1,397 toks

API · no tools. This is the prompt and response recorded for this attempt; a full solution may not have been returned.

Problem Statement

Return a single integer only. The problem has been revised so that no final answer is an irrational number, a parameterized expression, a range, a proof, a drawing, a graph, a locus equation, or a region description.

The original limit value is

L=\log 2-\frac12.

To avoid a logarithmic final answer, compute the integer

e^{L+1/2}.

Ground Truth

Recorded model response / solution

Reasoning Process

We are given that:

L = \log 2 - \frac{1}{2}

We are asked to compute:

e^{L + \frac{1}{2}}

Step 1: Substitute the expression for $L$

Substitute $L = \log 2 - $$\frac{1}{2}$$$ into the exponent:

L + \frac{1}{2} = \left( \log 2 - \frac{1}{2} \right) + \frac{1}{2} = \log 2

So the expression becomes:

e^{\log 2}

Step 2: Evaluate the exponential

Assuming that $\log$ refers to the natural logarithm (i.e., $\ln$ ), which is standard in calculus and mathematical analysis unless otherwise specified, we have:

e^{\log 2} = e^{\ln 2} = 2

This is because the exponential function $e^x$ and the natural logarithm $\ln x$ are inverse functions: $e^{\ln a} = a$ for any positive real number $a$ .

Final Answer

\boxed{2}

Final Model Output

Comparative Analysis (P15) (Average across runs)

Averages use reported measurements only. Missing usage is not treated as zero.

Avg Time Taken

K-EXAONE-236B-A23B

34.9

DeepSeek V4 Pro

5.0

Gemini 3.5 Flash

4.3

GPT-5.5

4.1

Claude Opus 4.8

2.7

Solar Pro 3

1.0

KT Mi:dm 2.0 Base Instruct

0.9

Avg Token Usage

K-EXAONE-236B-A23B

2002.0

KT Mi:dm 2.0 Base Instruct

901.0

Gemini 3.5 Flash

795.0

DeepSeek V4 Pro

367.7