TAOCP 1.2.2 Exercise 17
By definition (13), $\lg x = \log_2 x$.
Section 1.2.2: Numbers, Powers, and Logarithms
Exercise 17. ▶ [05] What is $\lg 32$? $\log_\pi \pi$? $\ln e$? $\log_b 1$? $\log_b(-1)$?
Verified: yes
Solve time: 33s
By definition (13), $\lg x = \log_2 x$. Therefore
$\lg 32 = \log_2 32 = 5,$
since $2^5 = 32$.
By definition (9), $\log_b(b) = 1$ for any $b \ne 1$, so
$\log_\pi \pi = 1, \qquad \ln e = \log_e e = 1.$
By the same definition, $\log_b 1 = 0$ for any $b \ne 1$, since $b^0 = 1$.
Finally, $\log_b(-1)$ is undefined for any real $b > 0$, $b \ne 1$, because no real exponent of a positive number produces a negative result.
Hence the values are
$\boxed{\lg 32 = 5, ;; \log_\pi \pi = 1, ;; \ln e = 1, ;; \log_b 1 = 0, ;; \log_b(-1) \text{ undefined}}.$