If A = log_2 log_2 log_4 256 + 2 log_{sqrt{2} | vedclass

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

Question

(C) Given $A = \log_2 \log_2 \log_4 256 + 2 \log_{\sqrt{2}} 2$. First,evaluate $\log_4 256$: Since $256 = 4^4$,$\log_4 256 = 4$. Next,evaluate $\log_2

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given $A = \log_2 \log_2 \log_4 256 + 2 \log_{\sqrt{2}} 2$. First,evaluate $\log_4 256$: Since $256 = 4^4$,$\log_4 256 = 4$. Next,evaluate $\log_2 \log_2 4$: Since $\log_2 4 = 2$,we have $\log_2 2 = 1$. Now,evaluate $2 \log_{\sqrt{2}} 2$: Since $\sqrt{2} = 2^{1/2}$,$\log_{2^{1/2}} 2 = \frac{1}{1/2} \log_2 2 = 2 	imes 1 = 2$. Thus,$2 \log_{\sqrt{2}} 2 = 2 	imes 2 = 4$. Finally,$A = 1 + 4 = 5$.

If $A = \log_2 \log_2 \log_4 256 + 2 \log_{\sqrt{2}} 2$,then the value of $A$ is:

Similar Questions

The real root of the equation ${\log _4}\{ {\log _2}(\sqrt {x + 8} - \sqrt x )\} = 0$ is..........

For a given number $\alpha > 1$,which of the following is the correct ascending order?

If ${x^{\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}} = \sqrt{3}}$,then $x$ is:

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

$\sinh[\log(2+\sqrt{5})] + \cosh[\log(2+\sqrt{3})] = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module