Question

Ans. (A). Explanation needed.​

Answer 1

Given,

$\longrightarrow \log_{10}x+\log_{10}x^{\frac{1}{2}}+\log_{10}x^{\frac{1}{4}}+\,\dots\,=y$

Since $\log a^b=b\log a,$

$\longrightarrow \log_{10}x+\dfrac{1}{2}\log_{10}x+\dfrac{1}{4}\log_{10}x+\,\dots\,=y$

$\longrightarrow \log_{10}x\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\,\dots\,\right)=y$

The series $1+\dfrac{1}{2}+\dfrac{1}{4}+\,\dots$ is an infinite geometric series of first term $1$ and common ratio $\dfrac{1}{2}.$

Then,

$\longrightarrow\dfrac{\log_{10}x\cdot1}{1-\dfrac{1}{2}}=y$

$\longrightarrow2\log_{10}x=y\quad\quad\dots(1)$

Here $y>0$ since $x>1.$

Given,

$\longrightarrow\dfrac{1+3+5+\,\dots\,+(2y-1)}{4+7+10+\,\dots\,+(3y+1)}=\dfrac{20}{7\log_{10}x}$

In LHS each numerator and denominator represents arithmetic series.

We have, if $a_1,\ a_2,\ a_3,\,\dots\,,\ a_n$ are in AP,

$\longrightarrow a_1+a_2+a_3+\,\dots\,+a_n=\dfrac{n}{2}(a_1+a_n)$

$\longrightarrow a_1+a_2+a_3+\,\dots\,+a_n=\dfrac{1}{2}\left(\dfrac{a_n-a_1}{d}+1\right)\bigg(a_1+a_n\bigg)$

Then,

$\longrightarrow\dfrac{\dfrac{1}{2}\left(\dfrac{2y-1-1}{2}+1\right)\bigg(1+2y-1\bigg)}{\dfrac{1}{2}\bigg(\dfrac{3y+1-4}{3}+1\bigg)\bigg(4+3y+1\bigg)}=\dfrac{20}{7\log_{10}x}$

$\longrightarrow\dfrac{(y-1+1)2y}{(y-1+1)(3y+5)}=\dfrac{20}{7\log_{10}x}$

$\longrightarrow\dfrac{2y^2}{y(3y+5)}=\dfrac{20}{7\log_{10}x}$

$\longrightarrow\dfrac{y}{3y+5}=\dfrac{10}{7\log_{10}x}$

$\longrightarrow7y\log_{10}x-10(3y+5)=0$

$\longrightarrow14y\log_{10}x-60y-100=0$

From (1),

$\longrightarrow7y^2-60y-100=0$

$\longrightarrow7y^2-70y+10y-100=0$

$\longrightarrow7y(y-10)+10(y-10)=0$

$\longrightarrow(y-10)(7y+10)=0$

Since $y>0,$

$\longrightarrow y=10$

$\longrightarrow 2\log_{10}x=10$

$\longrightarrow \log_{10}x=5$

$\longrightarrow\underline{\underline{x=10^5}}$

Hence (A) is the answer.

Answer 2

Step-by-step explanation:

m

$\\ Given, \\ \\ </p><p></p><p>\longrightarrow \log_{10}x+\log_{10}x^{\frac{1}{2}}+\log_{10}x^{\frac{1}{4}}+\ \\ ,\dots\,=y⟶log10x+log10x21+log10x41+…=y \\ \\ </p><p></p><p>Since \log a^b=b\log a,logab=bloga, \\ </p><p></p><p>\longrightarrow \log_{10}x+\dfrac{1}{2}\log_{10}x+\dfrac{1}{4}\log_{10}x+\ \\ ,\dots\,=y⟶log10x+21log10x+41log10x+…=y \\ </p><p></p><p>\longrightarrow \log_{10}x(1+\dfrac{1}{2}+\dfrac{1}{4}+\,\ \\ dots\,)=y⟶log10x(1+21+41+…)=y \\ \\ </p><p></p><p>The series 1+\dfrac{1}{2}+\dfrac{1}{4}+\,\dots1+21+41+… is an infinite geometric series of first term 11 and common ratio \dfrac{1}{2}.21. \\ </p><p></p><p>Then, \\ </p><p></p><p>\longrightarrow\dfrac{\log_{10}x\cdot1}{1-\dfrac{1}{2}}=y⟶1−21log10x⋅1=y \\ </p><p></p><p>\longrightarrow2\log_{10}x=y\quad\quad\dots(1)⟶2log10x=y…(1) \\ </p><p></p><p>Here y > 0y>0 since x > 1.x>1. \\ </p><p></p><p>Given, \\ </p><p></p><p>\longrightarrow\dfrac{1+3+5+\,\dots\,+(2y-1)}{4+7+10+\,\dots\,+(3y+1)}= \\ \\ \dfrac{20}{7\log_{10}x}⟶4+7+10+…+(3y+1)1+3+5+…+(2y−1)=7log10x20 \\ </p><p></p><p>In LHS each numerator and denominator represents arithmetic series. \\ </p><p></p><p>We have, if a_1,\ a_2,\ a_3,\,\dots\,,\ a_na1, a2, a3,…, an are in AP, \\ </p><p>\longrightarrow a_1+a_2+a_3+\,\dots\,+a_n=\dfrac{n}{2}(a_1+a_n)⟶a1+a2+a3+…+an=2n(a1+an) \\ </p><p></p><p> \\ \longrightarrow a_1+a_2+a_3+\,\dots\,+a_n=\dfrac{1}{2}(\dfrac{a_n-a_1}{d}+1)\bigg(a_1+a_n\bigg)⟶a1+a2+a3+…+an=21(dan−a1+1)(a1+an) \\ </p><p></p><p>Then, \\ </p><p></p><p>\longrightarrow\dfrac{\dfrac{1}{2}(\dfrac{2y-1-1}{2}+1)\bigg(1+2y-1\bigg)}{\dfrac{1}{2}\bigg(\dfrac{3y+1-4}{3}+1\bigg)\bigg(4+3y+1\bigg)}=\dfrac{20}{7\log_{10}x}⟶21(33y+1−4+1)(4+3y+1)21(22y−1−1+1)(1+2y−1)=7log10x20</p><p></p><p>\longrightarrow\dfrac{(y-1+1)2y}{(y-1+1)(3y+5)}=\dfrac{20}{7\log_{10}x}⟶(y−1+1)(3y+5)(y−1+1)2y=7log10x20</p><p></p><p>\longrightarrow\dfrac{2y^2}{y(3y+5)}=\dfrac{20}{7\log_{10}x}⟶y(3y+5)2y2=7log10x20</p><p></p><p>\longrightarrow\dfrac{y}{3y+5}=\dfrac{10}{7\log_{10}x}⟶3y+5y=7log10x10</p><p></p><p>\longrightarrow7y\log_{10}x-10(3y+5)=0⟶7ylog10x−10(3y+5)=0</p><p></p><p>\longrightarrow14y\log_{10}x-60y-100=0⟶14ylog10x−60y−100=0</p><p></p><p>From (1),</p><p></p><p>\longrightarrow7y^2-60y-100=0⟶7y2−60y−100=0</p><p></p><p>\longrightarrow7y^2-70y+10y-100=0⟶7y2−70y+10y−100=0</p><p></p><p>\longrightarrow7y(y-10)+10(y-10)=0⟶7y(y−10)+10(y−10)=0</p><p></p><p>\longrightarrow(y-10)(7y+10)=0⟶(y−10)(7y+10)=0</p><p></p><p>Since y > 0,y>0,</p><p></p><p>\longrightarrow y=10⟶y=10</p><p></p><p>\longrightarrow 2\log_{10}x=10⟶2log10x=10</p><p></p><p>\longrightarrow \log_{10}x=5⟶log10x=5</p><p></p><p>\longrightarrow\underline{\underline{x=10^5}}⟶x=105</p><p></p><p>Hence (A) is the answe</p><p></p><p>$