How do you rewrite #log_5x# as a ratio of common logs and natural logs?

1 Answer
Mar 7, 2017

Answer:

In common logarithm #log_5 x=logx/log5#

and in natural logarithm #log_5 x=lnx/ln5#

Explanation:

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_5 x=logx/log5#

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

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

i.e. #log_5 x=lnx/ln5#