How do you solve #2\log _ { 0,5} ( \log _ { 2} x ) + \log _ { 2} ( \log _ { 2} x ) = - 1#?

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Answer 1 · 2016-11-12T14:24:29

Start by writing in the same base.

#(2log_2(x))/(log0.5) + (log_2(x))/(log2) = -1#

#(2log_2(x))/(log(2^(-1))) + (log_2(x))/(log2^1) = -1#

#(2log_2(x))/(-1log2) + (log_2(x))/(1log2) = -1#

#(2log_2(x))/(-1log2) +(-1log_2(x))/(-1log2) = -1#

#(log_2(x^2))/(-1log2) + (log_2(1/x))/(-1log2) = -1#

#log_2(x^2 xx 1/x) = -log(1/2)#

#log_2(x) = -log(1/2)#

#x = 2^(-log(1/2))#

#x~=1.232#

Hopefully this helps!