he argument must be strictly positive: $$\frac{3x-1}{2-x}>0$$ The critical points are $x=\frac{1}{3}$ (numerator zero) and $x=2$ (denominator zero). Testing the three regions: for $x<\frac{1}{3}$ the numerator is negative and denominator positive, giving a negative; for $\frac{1}{3}2$ the numerator is positive and denominator negative. So: $$\text{Domain}=\left(\frac{1}{3},\ 2\right)$$ **Step 2: Unpack the inequality.** The base is $2>1$, so the logarithm is increasing and the inequality direction is **preserved** when exponentiating: $$\frac{3x-1}{2-x}<2^1=2$$ Move everything to one side rather than cross-multiplying, since $2-x$ changes sign: $$\frac{3x-1}{2-x}-2<0 \implies \frac{3x-1-2(2-x)}{2-x}<0 \implies \frac{5x-5}{2-x}<0$$ **Step 3: Solve.** Factor and tidy the sign: $$\frac{5(x-1)}{-(x-2)}<0 \implies \frac{x-1}{x-2}>0$$ A quotient is positive when both parts share a sign, giving $x<1$ or $x>2$. **Step 4: Intersect with the domain - this is where the question is won or lost.** $$\left[(-\infty,1)\cup(2,\infty)\right]\cap\left(\frac{1}{3},2\right)=\left(\frac{1}{3},\ 1\right)$$ The branch $x>2$ solves the algebra perfectly but lies entirely outside the domain, where the logarithm does not exist at all. Discarding it is not optional bookkeeping: carry it through and the answer set is simply wrong, no matter how clean the algebra behind it looked. **Final Answer:** $x \in \left( \frac{1}{3}, 1 \right)$2b:T43f,**The Setup:** This problem requires calculating the combined number of digits of two large numbers with a shared exponent. We can determine the number of digits by sandwiching the values between sequential powers of $10$. **Step 1:** Define the digit counting function. Let $d_1$ be the number of digits in $2^{2024}$. Thus, $10^{d_1-1} < 2^{2024} < 10^{d_1}$. Let $d_2$ be the number of digits in $5^{2024}$. Thus, $10^{d_2-1} < 5^{2024} < 10^{d_2}$. **Step 2:** Multiply the two bounding inequalities. $$10^{d_1-1} \times 10^{d_2-1} < 2^{2024} \times 5^{2024}