Gujarati
8. Sequences and Series
easy

$x$ के किस मान के लिए  ${\log _a}x + {\log _{\sqrt a }}x + {\log _{3\sqrt a }}x + ......... + {\log _{a\sqrt a }}x = \frac{{a + 1}}{2}$ होगा

A

$x = a$

B

$x = {a^a}$

C

$x = {a^{ - 1/a}}$

D

$x = {a^{1/a}}$

Solution

(d) ${\log _a}x + 2{\log _a}x + ……. + a{\log _a}x = \frac{{a + 1}}{2}$

$ \Rightarrow $ ${\log _a}x(1 + 2 + …….. + a) = \frac{{a + 1}}{2}$

 $ \Rightarrow $  ${\log _a}x.\frac{{a(a + 1)}}{2} = \frac{{a + 1}}{2}$

$ \Rightarrow $ $x=a^{1/a}$.

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.