How do I solve #log_0.25[ (log_2 3)(log_3 4)]# ?

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Answer 1 · 2017-10-14T16:02:12

The answer is #-1/2#.

Start by rewrite #0.25# as #1/4#.

#=log_(1/4)[(log_2 3)(log_3 4)]#

We can now use the change of base formula, which states that #log_a n =logn/loga#.

#=log_(1/4) [ (log3)/(log2) * log4/log3]#

#=log_(1/4) (log4/log2)#

#=log_(1/4) (log(2^2)/log2)#

#=log_(1/4) ((2log2)/log2)#

#=log_(1/4) 2#

Once again we use the change of base formula.

#=log2/(log(1/4)#

#=log2/(log2^-2)#

#=log2/(-2log2)#

#=-1/2#

Hopefully this helps!