Solución 3 de Notación Científica

Veamos los otros dos ejemplos de los problemas de notación científica:

Solución:

Veamos la solución a estos incisos:

i) ${\displaystyle \left(1.56\times {10}^{15}\right)+\left(2.45\times {10}^{16}\right)}$

Igualando bases:

${\displaystyle \begin{array}{l}\left(1.56\times {10}^{15}\right)+\left(2.45\times {10}^{16}\right)=\left(1.56\times {10}^{15}\right)+\left(24.5\times {10}^{15}\right)\\ \left(1.56\times {10}^{15}\right)+\left(2.45\times {10}^{16}\right)=\left(1.56+24.5\right)\times {10}^{15}\\ \left(1.56\times {10}^{15}\right)+\left(2.45\times {10}^{16}\right)=26.06\times {10}^{15}\\ \left(1.56\times {10}^{15}\right)+\left(2.45\times {10}^{16}\right)=2.606\times {10}^{16}\end{array}}$

j) ${\displaystyle \left(3.8\times {10}^9\right)\cdot \left(3.8\times {10}^{-9}\right)}$

Realizando la multiplicación:

${\displaystyle \begin{array}{l}\left(3.8\times {10}^9\right)\cdot \left(3.8\times {10}^{-9}\right)=\left(3.8\cdot 3.8\right)\times \left({10}^8\cdot {10}^{-9}\right)\\ \left(3.8\times {10}^9\right)\cdot \left(3.8\times {10}^{-9}\right)=14.44\times {10}^{9+\left(-9\right)}\\ \left(3.8\times {10}^9\right)\cdot \left(3.8\times {10}^{-9}\right)=14.44\times {10}^0\\ \left(3.8\times {10}^9\right)\cdot \left(3.8\times {10}^{-9}\right)=14.44\end{array}}$

¿Cómo te parecieron los resultados? ¿llegaste a los mismos?