If for $x, y \in \mathbb{R}, x>0, y = \log_{10} x + \log_{10} x^{1/3} + \log_{10} x^{1/9} + \dots$ upto $\infty$ terms and $\frac{2 + 4 + 6 + \dots + 2y}{3 + 6 + 9 + \dots + 3y} = \frac{4}{\log_{10} x}$, then the ordered pair $(x, y)$ is equal to :
Show Hint
In the fraction $\frac{\sum 2i}{\sum 3i}$, the variables cancel out to give exactly $2/3$ regardless of $y$. This quickly simplifies the second equation.
Step 1: Understanding the Concept:
The problem involves properties of logarithms, infinite Geometric Progression (GP), and Sum of Arithmetic Progression (AP). Step 2: Detailed Explanation:
1. Simplifying y:
\[ y = \log_{10} x (1 + 1/3 + 1/9 + \dots) \]
This is an infinite GP with $a=1, r=1/3$. Sum $S = \frac{1}{1-1/3} = \frac{3}{2}$.
So, $y = \frac{3}{2} \log_{10} x \implies \log_{10} x = \frac{2y}{3}$.
2. Simplifying the Ratio:
Numerator: $2(1 + 2 + \dots + y) = 2 \cdot \frac{y(y+1)}{2} = y(y+1)$.
Denominator: $3(1 + 2 + \dots + y) = 3 \cdot \frac{y(y+1)}{2}$.
Ratio = $\frac{y(y+1)}{\frac{3}{2}y(y+1)} = \frac{2}{3}$.
Given: $\frac{2}{3} = \frac{4}{\log_{10} x} \implies \log_{10} x = 6$.
3. Solving for x and y:
$\log_{10} x = 6 \implies x = 10^6$.
$y = \frac{3}{2} (6) = 9$. Step 3: Final Answer:
The ordered pair is $(10^6, 9)$.
