The Unnatural Log

use this formula to converted logs with different base $\\ \log_{ A}{ B} = \frac{ \log_{ P}{ B}}{ \log_{ Q}{ A}} \\$ (with P, Q are real numbers $\neq{ 1},>0 )\\$ So, $\\ \log_{ A}{ B} =\log_{ B}{ A} \\ \iff \frac{ \log_{ B}{ B}}{ \log_{ B}{ A}} =\log_{ B}{ A} \\ \iff \frac{ 1}{ \log_{ B}{ A}} =\log_{ B}{ A} \\ \iff (\log_{ B}{ A})^2 = 1 \\ \iff \log_{ B}{ A} = \pm{ 1} \\$ (by definition log) $\\ \iff A= B^{ \pm 1} \\ \iff A=B, or A= \frac{1}{B} \\$ But, $A \neq B.$ It means $A= \frac{1} {B},$ $A B = \boxed {1}$

Note $\log_AB=\frac{\log_BB}{\log_BA}=\frac{1}{\log_BA}=\frac{1}{\log_AB}.$

Therefore, $\log_AB=\pm1$ . Since $A\neq B$ , $\log_AB=-1$ . Then $B=\frac{1}{A}$ , so . $A\cdot B=\boxed{1}$ .

If we let $log_{A}B = x$ , the ecuation becomes $x=\frac{1}{x}$ $=>x^{2}-1=0$ $=>(x+1)(x-1)=0$ So $x$ can be $1$ or $-1$ , but $A\ne B$ , therefore $x=log_{A}B=-1 => A=\frac{1}{B}$

log(0.1)/log(10) = log(10)/log(0.1) = 1(Ans.)

ONLY 1 is possible value