If A-log-2-log-2-log-4-256- - 2 log-2-2- then A is equal to 7-2-3-5-

The correct option is D 5 A=log_2 (log_2 (log_4 256)) + 2 log_2^1/22 ∵ 256 = 4^4 log_ a^m = m log_ a log_b^n a = 1/n nbsp;log_b a ⇒ A= log_2 (log_2 (lo ...

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Byju's Answer

Standard IX

Mathematics

Logarithms

If A=log_2 (l...

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

$A = l o g_{2} (l o g_{2} (l o g_{4} 256)) + 2 l o g_{2^{\frac{1}{2}}} 2$
$∵$ 256 = $4^{4}$
$l o g a^{m} = m l o g a$
$l o g_{b^{n}} a = \frac{1}{n} l o g_{b} a$

$\Rightarrow A = l o g_{2} (l o g_{2} (l o g_{4} 4^{4})) + 2 \times \frac{1}{(\frac{1}{2})} l o g_{2} 2$
$= l o g_{2} (l o g_{2} 4) + 4 = l o g_{2} l o g_{2} 2^{2} + 4$
$= l o g_{2} 2 + 4 = 1 + 4 = 5$

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

The correct option is C 5 A=log2 (log2 (log4 256))+2log2122 ∵ 256 = 44 log am = m log a logbn a = 1nlogb a ⇒A=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