How do you rewrite #log_x(3/10)# as a ratio of common logs and natural logs?

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Answer 1 · 2017-11-24T05:18:39

#log(3/10)/logx=-0.5228/logx=-1.204/lnx#

We use the identity #log_ab=logb/loga# (common logs - base #10#) or #lnb/lna# (natural logs).

Hence converting #log_x(3/10)# in common logs, we have

#log(3/10)/logx#

= #(log3-log10)/logx#

= #(log3-1)/log#

= #(0.4771-1)/ogx=-0.5228/logx#

and converting #log_x(3/10)# to natural logs, we have

#ln(3/10)/lnx#

= #(ln0.3)/lnx#

= #-1.204/lnx#