If for x, y ∈ R, x > 0, y = log10^x + log10x^(1/3) + log10x^(1/9) + ..... upto ∞ terms and (2+4+6+...+2y)/(3+6+9+...+3y)=4/log10x

Question

If for x, y ∈ R, x > 0, y = log₁₀x + log₁₀x^1/3 + log₁₀x^1/9 + ..... upto ∞ terms and \(\frac{2+4+6+...+2y}{3+6+9+...+3y}\) = \(\frac{4}{log_{10}x}\),then the ordered pair (x, y) is equal to :

(2+4+6+...+2y)/(3+6+9+...+3y)=4/log₁₀x

(1) (10⁶,6) 

(2) (10⁴,6)

(3) (10²,3) 

(4) (10⁶,9)

Answer 1

Option : (4). (10⁶,9)

(2+4+6+...+2y)/(3+6+9+...+3y)=4/log₁₀x

\(\frac{2+4+6+...+2y}{3+6+9+...+3y}\) = \(\frac{4}{log_{10}x}\)

\(\frac{2(1+2+3+...+y)}{3(1+2+3+...+y)}\) = \(\frac{4}{log_{10}x}\)