# Nakafa Framework: LLM

URL: https://nakafa.com/en/exercises/high-school/snbt/quantitative-knowledge/try-out/set-7/11

Exercises: Try Out - Set 7: Real exam simulation to sharpen your skills and build confidence. - Problem 11

---

## Exercise 11

### Question

export const metadata = {
  title: "Problem 11",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

Given <InlineMath math="0 < x < 1" />. Which of the following relationships is correct between quantities <InlineMath math="P" /> and <InlineMath math="Q" /> based on the information provided?

| <InlineMath math="P" /> | <InlineMath math="Q" /> |
| :---: | :---: |
| <InlineMath math="\frac{4^x - 1}{2^x - 1}" /> | <InlineMath math="\frac{2^x + 1}{2^x}" /> |


### Choices

- [x] $$P > Q$$
- [ ] $$P < Q$$
- [ ] $$P = Q$$
- [ ] $$PQ = 1$$
- [ ] Cannot be determined

### Answer & Explanation

export const metadata = {
  title: "Solution for Problem 11",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "12/25/2025",
};

We will simplify the form of <InlineMath math="P" /> first.
Recall that <InlineMath math="4^x = (2^2)^x = (2^x)^2" />. Thus, the form <InlineMath math="4^x - 1" /> is a difference of squares which can be factored into <InlineMath math="(2^x - 1)(2^x + 1)" />.

<MathContainer>
  <BlockMath math="P = \frac{4^x - 1}{2^x - 1}" />
  <BlockMath math="P = \frac{(2^x - 1)(2^x + 1)}{2^x - 1}" />
  <BlockMath math="P = 2^x + 1" />
</MathContainer>

Next, let's look at the form of <InlineMath math="Q" />:
<MathContainer>
  <BlockMath math="Q = \frac{2^x + 1}{2^x}" />
</MathContainer>

We can see the relationship between <InlineMath math="P" /> and <InlineMath math="Q" /> by substituting <InlineMath math="P = 2^x + 1" /> into the equation for <InlineMath math="Q" />:
<MathContainer>
  <BlockMath math="Q = \frac{P}{2^x}" />
  <BlockMath math="P = Q \cdot 2^x" />
</MathContainer>

Given <InlineMath math="0 < x < 1" />. Let's analyze the value of <InlineMath math="2^x" /> in that range:
- If <InlineMath math="x > 0" />, then <InlineMath math="2^x > 2^0 \implies 2^x > 1" />.
- If <InlineMath math="x < 1" />, then <InlineMath math="2^x < 2^1 \implies 2^x < 2" />.

Thus, <InlineMath math="1 < 2^x < 2" />.

Back to the equation <InlineMath math="P = Q \cdot 2^x" />:
Since <InlineMath math="2^x > 1" /> and <InlineMath math="Q" /> is positive (because <InlineMath math="2^x > 0" />), multiplying <InlineMath math="Q" /> by a number greater than <InlineMath math="1" /> will result in a value greater than <InlineMath math="Q" /> itself.

<MathContainer>
  <BlockMath math="P = Q \cdot (\text{number} > 1)" />
  <BlockMath math="P > Q" />
</MathContainer>

Therefore, the correct relationship is <InlineMath math="P > Q" />.


---
