Why #log_(2)^2x(5/(log_(2)3*log_(2)4)+1/log_(2)^(2)5)=0# equal to #log_(2)^2x=0#?

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Answer 1 · 2018-06-30T21:49:51

I tried this:

The entire bit inside the brackets is a number, say, #n#; it can go to the right of the #=# sign as:

#log_2^2x=0/n=0#

giving your conclusion in the question;

we can continue and rearrange to write:

#sqrt(log_2^2x)=sqrt(0)#

#log_2x=0#

using the definitio of log:

#x=2^0=1#