Question

If A = log base 2 log base 2 log base 4 256+ 2 log base root 2 2. A=?

Answer 1

$a = log_{2}(?) log_{2}(?) log_{4}(256) + 2 log_{22}(?) \\ log_{2}(?) log_{2}(?) log_{4}(? {4}^{5} ) + 2 log_{ \sqrt{22} }(?) \\ log_{2}(?) log_{2}(?) 4 log_{4}(4) + 2 log_{ \sqrt{22} }(?) \\ log_{2}(?) log_{2}( {2}^{2} ) + 2 log_{ \sqrt{22} }(?) \\ log_{2}(2) + 2 log_{ \sqrt{22} }(?) \\ log_{2}(2) + 2 \times 1 \div 2 log_{22}(?) \\ log_{2}(22 \times \\ log_{2}(2) + log_{22}(1) \\ 1 + 0 = \\ = 1$

Answer 2

The value of A is 5.

Step-by-step explanation:

The given equation is

$A=\log_2(\log_2(\log_4(256)))+2\log_{\sqrt{2}}(2)$

It can be rewritten as

$A=\log_2(\log_2(\log_4(4^4)))+2\log_{\sqrt{2}}(\sqrt{2}^2)$

Using the properties of logarithm we get

$A=\log_2(\log_2(4))+2(2)$      $[\because \log_aa^x=x]$

$A=\log_2(\log_2(2^2))+4$

$A=\log_2(2)+4$

$A=1+4$

$A=5$

Therefore, the value of A is 5.

#Learn more

Properties of log.

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