▶️ Ejercicio 302 álgebra - Solucionario Baldor ✅✅

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CAPITULO XXXVIII

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Ejercicio 302

Hallar el número de términos de las progresiones:

$\begin{matrix} \div \div 3 & : 6 : ……… : 48 \\ S e a : \\ u = 48 \\ a = 3 \\ r = 2 \\ n & = \frac{l o g u - l o g a}{l o g r} + 1 \\ n & = \frac{l o g 48 - l o g 3}{l o g 2} + 1 \\ n & = \frac{1.681241237 - 0.4771212547}{0.3010299957} + 1 \\ n & = 4 + 1 \\ n & = 5 \end{matrix}$
$\begin{matrix} \div \div 2 & : 3 : ……… : \frac{243}{16} \\ S e a : \\ u = \frac{243}{16} \\ a = 2 \\ r = \frac{3}{2} \\ n = \frac{l o g u - l o g a}{l o g r} + 1 \\ n = \frac{l o g \frac{243}{16} - l o g 2}{l o g \frac{3}{2}} + 1 \\ n = \frac{l o g 243 - l o g 16 - l o g 2}{l o g 3 - l o g 2} + 1 \\ n = \frac{2.385606274 - 1.204119983 - 0.3010299957}{0.4771212547 - 1.204119983} + 1 \\ n = 5 + 1 \\ n = 6 \end{matrix}$
$\begin{matrix} \div \div 4 & : 8 : ……… : 512 \\ S e a : \\ u = 512 \\ a = 4 \\ r = 2 \\ n & = \frac{l o g u - l o g a}{l o g r} + 1 \\ n & = \frac{l o g 512 - l o g 4}{l o g 2} + 1 \\ n & = \frac{2.709269961 - 0.6020599913}{0.3010299957} + 1 \\ n & = 7 + 1 \\ n & = 8 \end{matrix}$
$\begin{matrix} \div \div 6 & : 8 : ……… : \frac{2048}{81} \\ S e a : \\ u = \frac{2048}{81} \\ a = 6 \\ r = \frac{4}{3} \\ n = \frac{l o g u - l o g a}{l o g r} + 1 \\ n = \frac{l o g (\frac{2048}{81}) - l o g 6}{l o g (\frac{4}{3})} + 1 \\ n = \frac{l o g 2048 - l o g 81 - l o g 6}{l o g 4 - l o g 3} + 1 \\ n = \frac{3.311329952 - 1.908485019 - 0.7781512504}{0.6020599913 - 0.4771212547} + 1 \\ n = 5 + 1 \\ n = 6 \end{matrix}$
$\begin{matrix} \div \div 2 & : 5 : ……… : \frac{625}{8} \\ S e a : \\ u = \frac{625}{8} \\ a = 2 \\ r = \frac{5}{2} \\ n = \frac{l o g u - l o g a}{l o g r} + 1 \\ n = \frac{l o g (\frac{625}{8}) - l o g 2}{l o g (\frac{5}{2})} + 1 \\ n = \frac{l o g 625 - l o g 8 - l o g 2}{l o g 5 - l o g 2} + 1 \\ n = \frac{2.795880017 - 0.903089987 - 0.3010299957}{0.6989700043 - 0.3010299957} + 1 \\ n = 4 + 1 \\ n = 5 \end{matrix}$

Categories: Capítulo XXXVIII