If for x, y in mathbb{R}, x 0, y = log_{10} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the expression for $y$:$y = (\log_{10} x) (1 + \frac{1}{3} + \frac{1}{9} + \dots \infty)$Using the sum of an infinite geometric series $S =

Vedclass · Accepted Answer

$(10^6, 9)$

Vedclass · Answer

(D) Given the expression for $y$:$y = (\log_{10} x) (1 + \frac{1}{3} + \frac{1}{9} + \dots \infty)$Using the sum of an infinite geometric series $S = \frac{a}{1-r}$ where $a=1$ and $r=\frac{1}{3}$:$y = (\log_{10} x) \left( \frac{1}{1 - 1/3} ight) = (\log_{10} x) \left( \frac{3}{2} ight) = \frac{3}{2} \log_{10} x$Now,consider the given equation:$\frac{2(1+2+3+\dots+y)}{3(1+2+3+\dots+y)} = \frac{4}{\log_{10} x}$$\frac{2}{3} = \frac{4}{\log_{10} x}$$\log_{10} x = \frac{4 	imes 3}{2} = 6$$x = 10^6$Substitute $\log_{10} x = 6$ into the expression for $y$:$y = \frac{3}{2} (6) = 9$Thus,the ordered pair $(x, y)$ is $(10^6, 9)$.

If for $x, y \in \mathbb{R}, x > 0,$ $y = \log_{10} x + \log_{10} x^{1/3} + \log_{10} x^{1/9} + \dots$ up to $\infty$ terms and $\frac{2+4+6+\dots+2y}{3+6+9+\dots+3y} = \frac{4}{\log_{10} x}$,then the ordered pair $(x, y)$ is equal to:

Similar Questions

Let $S$ be the sum of the first $9$ terms of the series: $(x+ka) + (x^2+(k+2)a) + (x^3+(k+4)a) + (x^4+(k+6)a) + \ldots$ where $a \neq 0$ and $x \neq 1$. If $S = \frac{x^{10}-x+45a(x-1)}{x-1}$,then $k$ is equal to:

If $a, b, c$ are in $G.P.$,and $a - b, c - a, b - c$ are in $H.P.$,then $a + 4b + c$ is equal to

If $9, x, y, z, a$ are in $A.P.$ such that $x + y + z = 15$,and $9, x, y, z, a$ are in $H.P.$ such that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3}$,then the value of $a$ is:

Let $\alpha = \sum_{n=101}^{200} 2^n \sum_{k=101}^n \frac{1}{k !}$ and $b = \sum_{n=101}^{200} \frac{2^{201}-2^n}{n !}$. Then,$\frac{a}{b}$ is

If $a$ is the arithmetic mean of $b$ and $c$ and $G_1, G_2$ are the two geometric means between them,then $G_1^3 + G_2^3 = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module