# How to check rationality, irrationality and algebraicity in this case?

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 would a calculator help? Hint: if $\log_{10} 5 =\frac ab$ then $5^b=2^a\times 5^a$,
– lulu
Mar 26 '19 at 17:58
• (D) What is $x$? A negative number? Mar 26 '19 at 17:58
• @DietrichBurde, It is not specified in the option. I think $x$ belongs to the general domain of the logarithmic function.
– MrAP
Mar 26 '19 at 18:00
• 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.
– lulu
Mar 26 '19 at 18:18
• 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.
– 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 $$0 since $$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.
• I mean $\pi=\log_e e^\pi$ is not algebraic. Mar 26 '19 at 19:11