Question

Corona y puntos al que me resuelva este ejercicio pls

Answer 1

Ecuaciones exponenciales

Encunciado
$\bf 3^{2m+1}=12\qquad\qquad 3^{n-1}=5\qquad \qquad 3^{m+n}=?\\ \\ \\ \bf 3^{2m+1}=12\qquad Aplicamos\ logaritmos\\ \\\\ \bf Log_33^{2m+1}=Log_312\\\\\\ \bf {(2m+1)}*Log_33=Log_312\\\\\\ \bf {(2m+1)}*1=2,26\\\\\\ \bf 2m+1-1=2,26-1\\\\\\ 2m=1,26\\\\\\ m= 1,26:2 \quad\to\quad m=0,63$

$\bf 3^{n-1}=5\\ \\ \\ \bf 3^{n-1}=5\qquad Aplicamos\ logaritmos\\ \\\\ \bf Log_33^{n-1}=Log_35\\\\\\ \bf {(n-1)}*Log_33=Log_35\\\\\\ \bf {(n-1)}*1=1,46\\\\\\ \bf n-1+1=1,46+1\\\\\\ n=2,46$

$\bf m= 0,63\qquad n= 2,46\\ \\ \\ 3^{m+n}= 3^{0,63+2,46}= 3^{3}=27\\\\Entonces \\\\ 3^{m+n}= 27$

Espero que te sirva, salu2!!!!