Solucion 1 de Notación Científica

Veamos la solución a los siguientes ejercicios de notación científica. Recuerda que estos son los primeros cuatro ejercicios del tema.

Solución:

a) ${\displaystyle \left(4\times {10}^4\right)\cdot \left(5\times {10}^{-3}\right)}$

Realizando la multiplicación:

${\displaystyle \begin{array}{l}\left(4\times {10}^4\right)\cdot \left(5\times {10}^{-3}\right)=\left(4\times 5\right)\times \left({10}^4\cdot {10}^{-3}\right)\\ \left(4\times {10}^4\right)\cdot \left(5\times {10}^{-3}\right)=20\times {10}^1\\ \left(4\times {10}^4\right)\cdot \left(5\times {10}^{-3}\right)=2\times {10}^2\end{array}}$

b) ${\displaystyle \left(9.1\times {10}^7\right)-\left(4.8\times {10}^8\right)}$

Igualando bases:

${\displaystyle \begin{array}{l}\left(9.1\times {10}^7\right)-\left(4.8\times {10}^8\right)=\left(9.1\times {10}^7\right)-\left(48\times {10}^7\right)\\ \left(9.1\times {10}^7\right)-\left(4.8\times {10}^8\right)=\left(9.1-48\right)\times {10}^7\\ \left(9.1\times {10}^7\right)-\left(4.8\times {10}^8\right)=-38.9\times {10}^7\\ \left(9.1\times {10}^7\right)-\left(4.8\times {10}^8\right)=-3.89\times {10}^8\end{array}}$

c) ${\displaystyle \left(1.8\times {10}^{-8}\right)+\left(7.7\times {10}^{-9}\right)}$

Igualando bases:

${\displaystyle \left(1.8\times {10}^{-8}\right)+\left(7.7\times {10}^{-9}\right)=\left(18\times {10}^{-9}\right)+\left(7.7\times {10}^{-9}\right)}$

${\displaystyle \begin{array}{l}\left(1.8\times {10}^{-8}\right)+\left(7.7\times {10}^{-9}\right)=\left(18\times {10}^{-9}\right)+\left(7.7\times {10}^{-9}\right)\\ \left(1.8\times {10}^{-8}\right)+\left(7.7\times {10}^{-9}\right)=\left(18+7.7\right)\times {10}^{-9}\\ \left(1.8\times {10}^{-8}\right)+\left(7.7\times {10}^{-9}\right)=25.7\times {10}^{-9}\\ \left(1.8\times {10}^{-8}\right)+\left(7.7\times {10}^{-9}\right)=2.57\times {10}^{-8}\end{array}}$

d) ${\displaystyle \left(5.43\times {10}^2\right)-\left(6.17\times {10}^{-2}\right)}$

Realizando la multiplicación

${\displaystyle \begin{array}{l}\left(5.43\times {10}^2\right)-\left(6.17\times {10}^{-2}\right)=\left(5.43\cdot 6.17\right)\times \left({10}^2\times {10}^{-2}\right)\\ \left(5.43\times {10}^2\right)-\left(6.17\times {10}^{-2}\right)=33.5031\times {10}^{2+\left(-2\right)}\\ \left(5.43\times {10}^2\right)-\left(6.17\times {10}^{-2}\right)=33.5031\times {10}^0\\ \left(5.43\times {10}^2\right)-\left(6.17\times {10}^{-2}\right)=33.5031\end{array}}$

¿Obtuviste los mismos resultados?