Diagnostic Test : Algebra ( Tes Aljabar James Stewart Kalkulus)

1. 

$(a). (-3)^4 = (-3)(-3)(-3)(-3)$

               $=9(-3)(-3)$

               $=-27(-3)$

               $=81$

$(b). -3^4 = -(3.3.3.3)=-81$

$(c). 3^{-4} = \frac{1}{3^4} = (\frac{1}{3})(\frac{1}{3})(\frac{1}{3})(\frac{1}{3}) =\frac{1}{81}$

$(d). \frac{5^23}{5^21} = \frac{(5^{21})(5^2)}{5^{21}} = 5^2 = 25$

$(e). (\frac{2}{3})^{-2} = \frac{1}{(\frac{2}{3})^{-2}} = (\frac{3}{2})^{2} =(\frac{3}{2}) (\frac{3}{2}) = \frac{9}{4}$

$(f). 16^\frac{-3}{4} = 4^{{(2)}{(\frac{-3}{4}})} = 4^\frac{-3}{2} = \sqrt{4^{-3}}= \sqrt{\frac{1}{4^{3}}} = \sqrt{\frac{1}{64}} = \frac{1}{8} $

2.

$(a). \sqrt{200}-\sqrt{32} = \sqrt{(100)(2)}-\sqrt{(16)(2)} = 10\sqrt{2}-4\sqrt{2}=6\sqrt{2}$

$(b). (3{a^3}{b^3})(4{a}{b^2})^2 =  (3{a^3}{b^3})(16{a^2}{b^4}) = 48{a^5}{b^7}$

$(c). \left(\frac{3x^{\frac{3}{2}}y^3}{x^{2}y^{\frac{-1}{2}}}\right)^{-2}= \frac{1}{\left(\frac{3x^{\frac{3}{2}}y^3}{x^{2}y^{\frac{-1}{2}}}\right)^{2}}= \left(\frac{x^{2}y^{\frac{-1}{2}}}{3x^{\frac{3}{2}}y^3}\right)^{2}= \left(\frac{x^{2}y^{\frac{-1}{2}}}{3x^{\frac{3}{2}}y^3}\right) \left(\frac{x^{2}y^{\frac{-1}{2}}}{3x^{\frac{3}{2}}y^3}\right)=\frac{x^{2+2}y^{\frac{-1}{2}+\frac{-1}{2}}}{9x^{\frac{3}{2}+\frac{3}{2}}y^{3+3}}=\frac{x^{4}y^{-1}}{9x^{3}y^6}= \frac{x^{4-3}}{9y^{6+1}}= \frac{x}{9y^{7}}$