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Question : Solve for x and y


$logx + {logx}^{ \frac{1}{2} } + {logx}^{ \frac{1}{4} } + - - - - = y$

and

$\sf \dfrac{1 + 3 + 5 + - - + (2y - 1)}{4 + 7 + 10 + - -+ (3y + 1)} = \dfrac{20}{7logx}$
​

Answer 1

Step-by-step explanation:

y=10 and x=10^5

I got these ans yr

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:logx + log {\bigg(x\bigg) }^{\dfrac{1}{2} } + log {\bigg(x\bigg) }^{\dfrac{1}{4} } + - - - - = y$

We know,

$\green{ \boxed{ \sf{ \: log(x) \: + \: log(y) \: = \: log(xy) \: }}}$

So, using this

$\rm :\longmapsto\:log\bigg[x {\bigg(x\bigg) }^{\dfrac{1}{2} } {\bigg(x\bigg) }^{\dfrac{1}{4} } - - - - \bigg] = y$

$\rm :\longmapsto\:log {\bigg(x \bigg) }^{1 + \dfrac{1}{2} + \dfrac{1}{4} + - - - - } = y$

We know, Sum of infinite GP series is given by

$\green{ \boxed{ \sf{ \:S_ \infty \: = \: \frac{a}{1 - r} \: }}}$

So, using this, we get

$\rm :\longmapsto\:log {\bigg(x \bigg) }^{ \dfrac{1}{1 - \frac{1}{2} } } = y$

$\rm :\longmapsto\:log {\bigg(x \bigg) }^{ \dfrac{1}{ \frac{1}{2} } } = y$

$\rm :\longmapsto\:log {\bigg(x \bigg) }^{ 2} = y$

$\rm :\longmapsto\:2 \: logx \: = \: y \: - - - (1)$

Also, given that

$\rm :\longmapsto\:\dfrac{1 + 3 + 5 + - - + (2y - 1)}{4 + 7 + 10 + - -+ (3y + 1)} = \dfrac{20}{7logx}$

can be rewritten as,

$\rm :\longmapsto\:\dfrac{\displaystyle\sum_{n=1}^y \: (2n - 1)}{\displaystyle\sum_{n=1}^y \: (3n + 1)} = \dfrac{40}{7 \times 2logx}$

$\rm :\longmapsto\:\dfrac{2\displaystyle\sum_{y=1}^n \: y - \displaystyle\sum_{y=1}^n1}{3\displaystyle\sum_{y=1}^n \: y + \displaystyle\sum_{y=1}^n1} = \dfrac{40}{7y}$

$\red{\bigg \{ \because \: 2logx = y\bigg \}}$

We know that,

$\green{ \boxed{ \sf{ \:\displaystyle\sum_{y=1}^n \: y = \frac{n(n + 1)}{2}}}} \: \: and \: \: \green{ \boxed{ \sf{ \:\displaystyle\sum_{y=1}^n \: 1 \: = \: n}}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{2\dfrac{y(y + 1)}{2} - y}{3\dfrac{y(y + 1)}{2} + y} = \dfrac{40}{7y}$

$\rm :\longmapsto\:\dfrac{ {2y}^{2} + 2y - 2y}{ {3y}^{2} + 3y + 2y} = \dfrac{40}{7y}$

$\rm :\longmapsto\:\dfrac{ {2y}^{2}}{ {3y}^{2} + 5y} = \dfrac{40}{7y}$

$\rm :\longmapsto\:\dfrac{ {2y}^{2}}{y( {3y} + 5)} = \dfrac{40}{7y}$

$\rm :\longmapsto\:\dfrac{ {2y}}{{3y} + 5} = \dfrac{40}{7y}$

$\rm :\longmapsto\:\dfrac{ {y}}{{3y} + 5} = \dfrac{20}{7y}$

$\rm :\longmapsto\: {7y}^{2} = 60y + 100$

$\rm :\longmapsto\: {7y}^{2} - 60y - 100 = 0$

$\rm :\longmapsto\: {7y}^{2} - 70y + 10y - 100 = 0$

$\rm :\longmapsto\:(y - 10)(7y + 10) = 0$

$\bf\implies \:y = 10 \: \: \: or \: \: \: \: y = - \: \dfrac{10}{7}$

So,

$\bf\implies \:2logx = 10 \: \: \: or \: \: \: \: 2logx = - \: \dfrac{10}{7}$

$\bf\implies \:logx = 5 \: \: \: or \: \: \: \: logx = - \: \dfrac{5}{7}$

We know,

$\green{ \boxed{ \sf{ \: log_{a}(b) = x \: \implies \: b \: = \: {a}^{x}}}}$

So, using this, we get

$\bf\implies \:x = {10}^{5} \: \: \: or \: \: \: x = {\bigg(10\bigg) }^{ - \dfrac{5}{7} }$