Rudin's defines the meaning of the symbol $x^{1/n}$ in theorem $1.21$
Theorem $1.21$ For every real $x>0$ and every integer $n>0$ there is one and only one positive real $y$ such that $y^n=x.$
This number $y$ is written $\sqrt[n]{x}$ or $x^{1/n}.$
Then in 6th problem of this chapter he first defines what the symbol $b^r$ means when $\color{red}{b>1}$ and $\color{red}r$ is a $\color{red}{\text{rational}}$ number. Then he proceeds to define $b^x$ where $\color{red}{b>1}$ and $\color{red}x$ is a $\color{red}{\text{real}}$ number.
- Fix $b>1.$
(a) If $m,n,p,q$ are integers, $n>0,q>0,$ and $r=m/n=p/q,$ prove that $$(b^m)^{1/n}=(b^p)^{1/q}.$$ Hence it makes sense to define $b^r=(b^m)^{1/n}.$
(b) Prove that $b^{r+s}=b^rb^s$ if $r$ and $s$ are rational.
(c) If $x$ is real, define $B(x)$ to be the set of all numbers $b^t,$ where $t$ is rational and $t\le x.$ Prove that $$b^r= \sup B(r)$$ when $r$ is rational. Hence it makes sense to define $b^x=\sup B(x)$ for every real $x.$
(d) Prove that $b^{x+y}=b^xb^y$ for all real $x$ and $y.$
Finally, on problem no. 7 Rudin defines the logarithm of $y$ to the base $b$ where $\color{red}{b>1}$ and $\color{red}y$ is a $\color{red}{\text{ positive real}}$ number.
- Fix $b > 1$, $y > 0$, and prove that there is a unique real number $x$ such that $b^x = y$, by completing the following outline. (This $x$ is called the logarithm of $y$ to the base $b$. )
Let $A$ be the set of all $w$ such that $b^w < y$, and show that $x = \sup A$ satisfies $b^x = y$. Prove that this $x$ is unique.
$\ldots$ and Rudin ends it here. But I think we all have used the log when $b$ is between $0$ and $1.$
Question : How does one extend this definition of logarithm to $b$ for $0<b<1$?(or maybe more?)
Best Answer
Once you have defined $\log_b x$ for $b\gt 1$, you can define $\log_b x =-\log_{1/b} x$ for $0\lt b \lt 1$