How do you evaluate #log_20 0.125# using the change of base formula?

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Answer 1 · 2018-04-15T16:14:22

#log_20 0.125=-0.6941#

According to change of base formula

#log_ab=log_xb/log_xa=logb/loga#

and when we use base #10#, we do not write #10# in subscript.

Hence #log_20 0.125#

= #log0.125/log20#

= #log(1/8)/log(2*10)#

= #(log1-log8)/(log2+log10)#

= #(-log2^3)/(log2+1)#

= #-(3log2)/(1+log2)#

= #-(3*0.3010)/1.3010#

= #-0.6941#