If for x, y [] R, x 0, y = log10x + log10x1/3 + log10x1/9 + ......

Question

If for x, y $$\in$$ R, x > 0, y = log10x + log10x1/3 + log10x1/9 + ...... upto $$\infty$$ terms

and $${{2 + 4 + 6 + .... + 2y} \over {3 + 6 + 9 + ..... + 3y}} = {4 \over {{{\log }_{10}}x}}$$, then the ordered pair (x, y) is equal to :

Accepted Answer

$${{2(1 + 2 + 3 + .... + y)} \over {3(1 + 2 + 3 + .... + y)}} = {4 \over {{{\log }_{10}}x}}$$

$$ \Rightarrow {\log _{10}}x = 6 \Rightarrow x = {10^6}$$

Now,

$$y = ({\log _{10}}x) + \left( {{{\log }_{10}}{x^{{1 \over 3}}}} \right) + \left( {{{\log }_{10}}{x^{{1 \over 9}}}} \right) + ....\infty $$

$$ = \left( {1 + {1 \over 3} + {1 \over 9} + ....\infty } \right){\log _{10}}x$$

$$ = \left( {{1 \over {1 - {1 \over 3}}}} \right){\log _{10}}x = 9$$

So, (x, y) = (106, 9)

Solutions · 1