Question

a

Answer 1

\(\bf log_{\color{red}{ a}}{\color{blue}{ b}}=y\implies {\color{red}{ a}}^y={\color{blue}{ b}}
\\ \quad \\ \quad \\
log_{\color{red}{ 2}}{\color{blue}{ 2}}=\square \implies ?\)

Answer 2

log2(2) = 1

Answer 3

logx^(logx)?

Answer 4

**log^(logx)x?

Answer 5

yeap thus \(\bf log_{\color{red}{ a}}{\color{blue}{ b}}=y\implies {\color{red}{ a}}^y={\color{blue}{ b}}
\\ \quad \\ \quad \\
log_{\color{red}{ 2}}{\color{blue}{ 2}}=1\implies {\color{red}{ 2}}^1={\color{blue}{ 2}}\)

so it simplifies to 1 :)