How do you rewrite #log_x(3/4)# as a ratio of common logs and natural logs?

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Answer 1 · 2017-03-03T12:26:53

In common logarithm #log_x (3/4)=log(3/4)/logx#

and in natural logarithm #log_x (3/4)=ln(3/4)/lnx#

Let #log_b a=x#, then we know that #b^x=a#

and taking common logs (i.e. to the base #10#) on both sides, we get

#logb^x=loga# i.e. #xlogb=loga# or #x=loga/logb#

As #x=log_b a#, we can say #log_b a=loga/logb#

Hence in common logarithm #log_x (3/4)=log(3/4)/logx#

Now as #b^x=a#, taking natural logarithm we get

#lnb^x=lna# i.e. #xlnb=lna# or #x=lna/lnb#

i.e. #log_x (3/4)=ln(3/4)/lnx#