"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 :"

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)"

If for x, y $\in$ R, x > 0, y = log₁₀x + log₁₀x^1/3 + log₁₀x^1/9 + ...... 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 :

Solution

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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 :

Solution

About this question

More Sequences and Series questions

Drill 25 more like these. Every day. Free.

If for x, y $\in$ R, x > 0, y = log₁₀x + log₁₀x^1/3 + log₁₀x^1/9 + ...... 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 :