How do you solve # log _(2) (log _(2) (log _(2) x))) = 2#?

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Answer 1 · 2015-12-17T10:05:31

Make repeated use of the fact that #2^(log_2(x)) = x# to find that
#x = 65536#

Applying the property of logarithms that

#a^(log_a(x)) = x#

we have

#log_2(log_2(log_2(x))) = 2#

#= 2^(log_2(log_2(log_2(x)))) = 2^2#

#=> log_2(log_2(x)) = 4#

#=> 2^(log_2(log_2(x))) = 2^4#

#=> log_2(x) = 16#

#=> 2^(log_2(x)) = 2^16#

#=> x = 65536#

Answer 2 · 2015-12-17T10:17:42

#x=2^16=65536#

Start from this point: if you know that #a=b#, then it must be #2^a=2^b#. So, start from your equation, and deduce that

#2^{log_2(log_2(log_2(x)))}=2^2#

Now, by definition, #2^{log_2(z)} = z#, while of course #2^2=4#. So, the equation becomes

#log_2(log_2(x))=4#

And now we're in the same situation as before, so we can apply the same logic:

#2^{log_2(log_2(x))}=2^4#

Which means

#log_2(x)=16#

One last iteration gives us

#x=2^16=65536#

Verify the answer:

#log_2(log_2(log_2(2^16))) = log_2(log_2(16))=log_2(4)=2#