Respuesta :
[tex]\bf log_{{ a}}(xy)\implies log_{{ a}}(x)+log_{{ a}}(y)
\\\\\\
% Logarithm of exponentials
log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)
\\\\\\
{{ a}}^{log_{{ a}}x}=x\impliedby \textit{log cancellation rule}\\\\
-----------------------------\\\\[/tex]
[tex]\bf log(x)=2.46-1.12log(y)\iff log_{10}(x)=2.46-1.12log_{10}(y) \\\\\\ log_{10}(x)=2.46-log_{10}(y^{1.12})\implies log_{10}(x)+log_{10}(y^{1.12})=2.46 \\\\\\ log_{10}(x\cdot y^{1.12})=2.46\implies 10^{\cfrac{}{}log_{10}(x\cdot y^{1.12})}=10^{2.46} \\\\\\ xy^{1.12}=10^{2.46}\implies \boxed{x=\cfrac{10^{2.46}}{y^{1.12}}}[/tex]
[tex]\bf log(x)=2.46-1.12log(y)\iff log_{10}(x)=2.46-1.12log_{10}(y) \\\\\\ log_{10}(x)=2.46-log_{10}(y^{1.12})\implies log_{10}(x)+log_{10}(y^{1.12})=2.46 \\\\\\ log_{10}(x\cdot y^{1.12})=2.46\implies 10^{\cfrac{}{}log_{10}(x\cdot y^{1.12})}=10^{2.46} \\\\\\ xy^{1.12}=10^{2.46}\implies \boxed{x=\cfrac{10^{2.46}}{y^{1.12}}}[/tex]
