Ecuaciones exponenciales Encunciado [tex]\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[/tex]
[tex]\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[/tex][tex]\bf m= 0,63\qquad n= 2,46\\ \\ \\ 3^{m+n}= 3^{0,63+2,46}= 3^{3}=27\\\\Entonces \\\\ 3^{m+n}= 27[/tex]
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Ecuaciones exponenciales
Encunciado
[tex]\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[/tex]
[tex]\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[/tex]
[tex]\bf m= 0,63\qquad n= 2,46\\ \\ \\ 3^{m+n}= 3^{0,63+2,46}= 3^{3}=27\\\\Entonces \\\\ 3^{m+n}= 27[/tex]
Espero que te sirva, salu2!!!!