If A -log 2log 2log 4 256-2 log -2 2- then A is equal toA- 7B- 2C- 3D- 5

The correct option is D 5A=log_2 log_2 log_4 256 + 2 log_2_1/22 = log_2 log_2 log_4 4^4+2 ×1/ ( 1/2 )log_2 2 =log_2 log_2 4+4 = log_2 log_2 2^2 + 4 =log_2 2+4= ...

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Byju's Answer

Standard IX

Mathematics

Logarithms

If A =log 2lo...

Question

If $A = l o g_{2} l o g_{2} l o g_{4} 256 + 2 l o g_{\sqrt{2}} 2$ , then A is equal to

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Solution

The correct option is C 5
$A = l o g_{2} l o g_{2} l o g_{4} 256 + 2 l o g_{2_{\frac{1}{2}}} 2$
$= l o g_{2} l o g_{2} l o g_{4} 4^{4} + 2 \times \frac{1}{(\frac{1}{2})} l o g_{2} 2$
$= l o g_{2} l o g_{2} 4 + 4 = l o g_{2} l o g_{2} 2^{2} + 4$
$= l o g_{2} 2 + 4 = 1 + 4 = 5$

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Similar questions

Q. If

A = l o g_{2} l o g_{2} l o g_{4} 256 + 2 l o g_{\sqrt{2}} 2

, then A is equal to

Q. If

A = {log}_{2} {log}_{2} {log}_{4} 256 + 2 {log}_{\sqrt{2}} 2

, then

A

is equal to

Q. The value of

{log}_{2} {log}_{2} {log}_{4} 256 + 2 {log}_{\sqrt{2}} 2

is equal to

l o g 2 l o g 2 l o g_{4} 256 + 2 l o g_{\sqrt{2}} 2

\lim_{n \to \infty} \frac{1^{2} + 2^{2} + 3^{2} + . . . + n^{2}}{n^{3}}

is equal to

(a) 1
(b) 1/2
(c) 1/3
(d) 0

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If A=log2 log2 log4 256+2log√22, then A is equal to

The correct option is C 5A=log2 log2 log4 256+2log2122 =log2 log2 log4 44+2×1(12)log22 =log2 log24+4=log2 log2 22+4 =log22+4=1+4=5

If $A = l o g_{2} l o g_{2} l o g_{4} 256 + 2 l o g_{\sqrt{2}} 2$ , then A is equal to

The correct option is C 5
$A = l o g_{2} l o g_{2} l o g_{4} 256 + 2 l o g_{2_{\frac{1}{2}}} 2$
$= l o g_{2} l o g_{2} l o g_{4} 4^{4} + 2 \times \frac{1}{(\frac{1}{2})} l o g_{2} 2$
$= l o g_{2} l o g_{2} 4 + 4 = l o g_{2} l o g_{2} 2^{2} + 4$
$= l o g_{2} 2 + 4 = 1 + 4 = 5$