is log(e) another way of writing ln ? as $$ log_{(e)} = ln $$

and consequently when x=1 $$ log_{(e)}(1+1/2x) = ln(1+1/2x) = ln(3/2) =0.405 $$

Please let me know.

Thank you.

  • $\begingroup$ So you did not see any thing wrong with above statements ? That means log(e) is ln right ? $\endgroup$ – tt z Jan 14 '19 at 16:28
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    $\begingroup$ $\log_e$ is the same as $\ln$. There's no need for brackets round the $e$. $\endgroup$ – timtfj Jan 14 '19 at 17:45
  • $\begingroup$ I don't think this is a duplicate of the linked question—this question is specifically about equivalence of notation, and the other question is more about the difference between logs to base $10$ and $e$. Its answers focus mainly on differences between the two functions. $\endgroup$ – timtfj Jan 14 '19 at 17:51