Question

1) Se x e y são números reais tais que 2ˣ = m e $2^y$ = n, então $4^{x-y}$ é igual a:

a) 2(m-n)
b) $\frac{m-n}{2}$
c) $- \frac{m}{n}$
d) $\frac{m^2}{n^2}$
e) $2^{ \frac{m}{n}$

2) O valor da expressão $4.(0,5)^4 + \sqrt{0,25} + 8^{- \frac{2}{3} }$ é:

a) 1/8
b) 1/4
c) 1/2
d) 1
e) 2

3) Se 2ˣ = a e $2^y$ = b, para x e y reais, então o valor de $(0,25)^{-3x+y}$ é igual a:

a) a⁶b²
b) b²a⁻⁶
c) a⁻⁶b⁻²
d) a⁶b⁻²
e) $\sqrt{a^6b^2}$

Answer 1

1)
$4^{x-y}=\frac{4^{x}}{4^{y}}=\frac{(2^{2})^{x}}{(2^{2})^{y}}=\frac{2^{2x}}{2^{2y}}=\frac{(2^{x})^{2}}{(2^{y})^{2}}=\frac{m^{2}}{n^{2}}$

2)
$4.(0,5)^{4}+\sqrt{0,25}+8^{-\frac{2}{3}}=\\\\=4.(5.10^{-1})^{4}+\sqrt{25.10^{-2}}+\frac{1}{8^{\frac{2}{3}}}=\\\\=4.625.10^{-4}+5.10^{-1}+\frac{1}{\sqrt[3]{8^{2}}}=\\\\=2500.10^{-4}+0,5+\frac{1}{\sqrt[3]{(2^{3})^{2}}}=\\\\=0,25+0,5+\frac{1}{\sqrt[3]{2^{6}}}=\\\\=0,75+\frac{1}{2^{2}}=\\\\=0,75+\frac{1}{4}=\\\\=\frac{3+1}{4}=\\\\=\frac{4}{4}=\\\\=1$

3)
$(0,25)^{-3x+y}=\\\\=(\frac{1}{4})^{-3x+y}=\\\\=(\frac{1}{4})^{-3x}.(\frac{1}{4})^{y}=\\\\=(\frac{1^{-3x}}{4^{-3x}}).(\frac{1^{y}}{4^{y}})=\\\\=4^{3x}.4^{-y}=\\\\=(2^{2})^{3x}.(2^{2})^{-y}=\\\\=2^{6x}.2^{-2y}=\\\\=(2^{6})^{x}.(2^{-2})^{y}=\\\\=(2^{x})^{6}.(2^{y})^{-2}=\\\\=a^{6}.b^{-2}$