Respuesta :
[tex]\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log_0(3)=x\implies 0^x=3~\hfill \stackrel{\textit{so there's no \underline{x} value that will satisfy that}}{0^{anything}=0\qquad \qquad 0^0=unde fined} \\\\\\ \log_1(3)=x\implies 1^x=3\hfill \stackrel{\textit{so there's no \underline{x} value that will satisfy that either}}{1^{anything}=1}[/tex]
Answer:
See below.
Step-by-step explanation:
By the definition of a logarithm:-
If y = log0 3 then 3 = 0^y = 0 which is obviously wrong.
If y = log1 3 then 3 = 1 ^y = 1 which again is wrong.
Neither 0 nor 1 can be a base of logarithms. 0 to any power will equal 0 and 1 to any power equals 1.
