Question

Tentukan hasil pemangkatan bilangan bilangan berikut
[tex]a. \: {49}^{ \frac{1}{2} } [/tex]
[tex]b. \: { - 8}^{ \frac{2}{3} } [/tex]
[tex]c. \: ( - { \frac{8}{7} }) ^{ \frac{1}{3} } [/tex]
[tex]d. \: ( \frac{1}{ {16x}^{4} } ) ^{ \frac{1}{4} } [/tex]

​

Answer 1

Jawab:

Penjelasan dengan langkah-langkah:

a. 49^½ = (7²)^½

= 7^(2 × ½)

= 7

b. -8^⅔ = (-(2)³)^⅔

= - (2)^(3 × ⅔)

= - (2)²

= - 4

c. (-8/7)^⅓ = ((-8)^⅓)/(7^⅓)

= (((-2)³)^⅓)/(7^⅓)

= ((-2)^(3×⅓))/(7^⅓)

= -2/(7^⅓)

= -2/³V7

d. (1/(16x⁴))^¼ = (16x⁴)^(-¼)

= ((2x)⁴)^(-¼)

= (2x)^(4 × (-¼))

= (2x)^(-1)

= 1/(2x)

^ = pangkat

V = akar

Answer 2

Verified answer

Jawab:
[tex]\displaystyle(a.)\:\:=7\\(b.)\:\:=-4\\\\(c.)\:\:=-\frac{2(\sqrt[3]{49})}{7} \\\\(d.)\:\:=\frac{1}{2x}\:\left\{ x\ne0\right\}[/tex]

Penjelasan dengan langkah-langkah:

[tex]\displaystyle\because a^{\frac{b}{c}}=\sqrt[c]{a^b} \therefore\\\\(a.)\:\:49^{\frac{1}{2}}=\sqrt[2]{49^1}=\sqrt{49^1}=\sqrt{49}=7\\\\(b.)\:\:-8^{\frac{2}{3}}=-(2^{\not3})^{\frac{2}{\not3}}=-(2^2)=-4[/tex]

[tex]\displaystyle (c.)\:\:\left(-\frac{8}{7}\right)^{\frac{1}{3}}=-\left(\frac{8}{7}\right)^{\frac{1}{3}}=-\sqrt[3]{\left(\frac{8}{7}\right)^1} \\~=-\sqrt[3]{\frac{8}{7}} \:\:\because\sqrt[c]{\frac{a}{b}} =\frac{\sqrt[c]{a}}{\sqrt[c]{b}} \therefore-\sqrt[3]{\frac{8}{7}} =-\frac{\sqrt[3]{8}}{\sqrt[3]{7}}\\\\~=-\frac{2}{\sqrt[3]{7}}=-\frac{2\sqrt[3]{7}\sqrt[3]{7}}{\sqrt[3]{7}\sqrt[3]{7}\sqrt[3]{7}}=-\frac{2(\sqrt[3]{7})^2}{7}\\~=-\frac{2(\sqrt[3]{7^2})}{7}~=-\frac{2(\sqrt[3]{49})}{7}[/tex]

[tex]\displaystyle(d.)\:\:\left(\frac{1}{16x^4}\right)^{\frac{1}{4}}=\left(\frac{1}{(2^4)x^4}\right)^{\frac{1}{4}}=\left(\frac{1}{(2x)^4}\right)^{\frac{1}{4}}\\\\\because\frac{1}{a^n}=a^{-n}\therefore\left(\frac{1}{(2x)^4}\right)^{\frac{1}{4}}=(2x^{-\not4})^{\frac{1}{\not4}}\\\\=(2x)^{-1}=\frac{1}{2x}\:\left\{ x\ne0\right\}[/tex]

(xcvi)