Question

é sobre função exponencial, dou 20 pontos pra que responder a questão no anexo

Answer 1

$\frac{2^{-3}.0,5^4.8}{\left(\frac{1}{2}\right)^3.2^{-2}.\left(\frac{1}{4}\right)^{-2}}\\\\\\\frac{2^{-3}.\left(2^{-1}\right)^4.2^3}{\left(2^{-1}\right)^3.2^{-2}.\left(\frac{1^2}{2^2}\right)^{-2}}\\\\\\\frac{2^{-3}.\left(2\right)^{-1\;.\;4}.2^3}{\left(2\right)^{-1\;.\;3}.2^{-2}.\left(\frac{1}{2}\right)^{\;2\;.\;-2}}$

$\\\frac{2^{-3}.\left(2\right)^{-1\;.\;4}.2^3}{\left(2\right)^{-1\;.\;3}.2^{-2}.\left(2^{-1}\right)^{-4}}\\\\\\\frac{2^{-3}.\left(2\right)^{-1\;.\;4}.2^3}{\left(2\right)^{-1\;.\;3}.2^{-2}.\left(2\right)^{-1\;.\;-4}}\\\\\\\frac{2^{-3}.\left(2\right)^{-4}.2^3}{\left(2\right)^{-3}.2^{-2}.\left(2\right)^{4}}\\\\\\\frac{2^{-3+(-4)+3}}{2^{-3+(-2)+4}}\\\\\\\frac{2^{-4}}{2^{-1}}\\\\\\2^{-4-(-1)}\\\\2^{-4+1}\\\\2^{-3}\\\\\frac{1^3}{2^3}\\\\\frac{1}{8}\\\\0,125$