This question was asked in a test and the use of calculator was not allowed.

Choose the correct one:

(A) $\log_e x$ can be defined as a real-valued function of $x$ for all $x\in R$

(B) $\log_{10}5$ is a rational number

(C) $\log_{10}5$ is an irrational number

(D) $\log_e x$ is algebraic number

I know that (A) cannot be the answer because $\log_e x$ is not defined for negative values of $x$

How do I check the validity of the other options without using a calculator. Please help.

  • $\begingroup$ How would a calculator help? Hint: if $\log_{10} 5 =\frac ab$ then $5^b=2^a\times 5^a$, $\endgroup$
    – lulu
    Mar 26 '19 at 17:58
  • 1
    $\begingroup$ (D) What is $x$? A negative number? $\endgroup$ Mar 26 '19 at 17:58
  • $\begingroup$ @DietrichBurde, It is not specified in the option. I think $x$ belongs to the general domain of the logarithmic function. $\endgroup$
    – MrAP
    Mar 26 '19 at 18:00
  • 1
    $\begingroup$ Well...that wouldn't prove anything. An apparent pattern might not persist and the absence of a pattern might only mean that the period is bigger than the output of your calculator. $\endgroup$
    – lulu
    Mar 26 '19 at 18:18
  • 1
    $\begingroup$ For instance, in this case we have $\log_{10}5=0.6989700043360188047\cdots$. That certainly looks irrational but how do you know that it doesn't eventually become periodic? Numerical methods really don't help with questions like these. $\endgroup$
    – lulu
    Mar 26 '19 at 18:20

Assume by contrary that $\log_{10}5\in \Bbb Q$ then we have$$\log_{10}5={p\over q}\quad,\quad \gcd(p,q)=1$$with $0<p<q$ since $1<5<10$, therefore$$5=10^{p\over q}\implies 5^q=10^p\implies 5^{q-p}=2^p$$which is impossible since no power of $2$ can be any power on $5$ except $1=2^0=5^0$. Finally we conclude that $\log_{10}5\notin \Bbb Q$ and (C) is correct.

Also (D) is wrong. Let $x=e^\pi$ and note that $\pi$ is transcendental.

  • $\begingroup$ I could not understand the last line of your answer. $\endgroup$
    – MrAP
    Mar 26 '19 at 19:09
  • $\begingroup$ I mean $\pi=\log_e e^\pi$ is not algebraic. $\endgroup$ Mar 26 '19 at 19:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.