The solution of log_{sqrt{3}} x + log_{sqrt[4 | vedclass

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

Question

(D) Given equation: $\log_{\sqrt{3}} x + \log_{\sqrt[4]{3}} x + \log_{\sqrt[6]{3}} x + \dots + \log_{\sqrt[16]{3}} x = 36$Using the property $\log_{a^

Vedclass · Accepted Answer

$x = \sqrt{3}$

Vedclass · Answer

(D) Given equation: $\log_{\sqrt{3}} x + \log_{\sqrt[4]{3}} x + \log_{\sqrt[6]{3}} x + \dots + \log_{\sqrt[16]{3}} x = 36$Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$,we have $\log_{3^{1/k}} x = k \log_3 x$.Substituting this into the equation:$2 \log_3 x + 4 \log_3 x + 6 \log_3 x + \dots + 16 \log_3 x = 36$Factor out $\log_3 x$:$(\log_3 x) (2 + 4 + 6 + \dots + 16) = 36$The sum of the arithmetic progression $2 + 4 + \dots + 16$ with $n = 8$ terms is $\frac{8}{2}(2 + 16) = 4 	imes 18 = 72$.So,$(\log_3 x) 	imes 72 = 36$$\log_3 x = \frac{36}{72} = \frac{1}{2}$$x = 3^{1/2} = \sqrt{3}$.

The solution of $\log_{\sqrt{3}} x + \log_{\sqrt[4]{3}} x + \log_{\sqrt[6]{3}} x + \dots + \log_{\sqrt[16]{3}} x = 36$ is

Similar Questions

The set of real values of $x$ for which $\log_{0.2} \left( \frac{x + 2}{x} \right) \le 1$ is

If $x+\log _{10}\left(1+2^{x}\right)=x \log _{10} 5+\log _{10} 6$, then the value of $x$ is

For $y = \log_a x$ to be defined,$a$ must be:

The number $\log_{20} 3$ lies in

If $x$ is a positive real number different from $1$ such that $\log _{a} x, \log _{b} x, \log _{c} x$ are in $AP$, then

Vedclass Test Series

Exam Paper Generator

Online Exam Module