Question

dengan cara ya coy, cuma 4 doang dikit
yang ngasal dici pok fuad​

Answer 1

Verified answer

Nomor 1

[tex] \frac{( {x}^{n - 1} {y}^{n}) {}^{3} }{ {x}^{2n} {y}^{ 6 + n} } = {x}^{a} {y}^{b} [/tex]

[tex] \frac{ {x}^{3 n - 3} {y}^{3n} }{ {x}^{2n} {y}^{6 + n} } = {x}^{a} {y}^{b} [/tex]

[tex] {x}^{(3n - 3) - 2n} \: {y}^{3n - (6 + n)} = {x}^{a} {y}^{b} [/tex]

[tex] {x}^{n - 3} \: {y}^{2n - 6} = {x}^{a} {y}^{b} [/tex]

[tex]a = n - 3[/tex]

[tex]b = 2n - 6[/tex]

maka nilai dari :

[tex] = \frac{b}{a} [/tex]

[tex] = \frac{2n - 6}{n - 3} [/tex]

[tex] = \frac{2(n - 3)}{n - 3} [/tex]

[tex] = \frac{2 \cancel{(n - 3)}}{ \cancel{n - 3}} [/tex]

[tex] = 2[/tex]

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Nomor 2

a

[tex] {3}^{n} = 243[/tex]

[tex] {3}^{n} = {3}^{5} [/tex]

[tex]n = 5[/tex]

b

[tex] {2}^{n + 1} = \frac{1}{16} [/tex]

[tex] {2}^{n + 1} = \frac{1}{ {2}^{4} }[/tex]

[tex] {2}^{n + 1} = {2}^{ - 4} [/tex]

[tex]n + 1 = - 4[/tex]

[tex]n = - 5[/tex]

c

[tex] {4}^{n} = ( - 2) {}^{0} [/tex]

[tex]( {2}^{2} ) {}^{n} = 1[/tex]

[tex] {2}^{2n} = {2}^{0} [/tex]

[tex]2n = 0[/tex]

[tex]n = 0[/tex]

d

[tex]48 \div 3 = {n}^{4} [/tex]

[tex]16 = {n}^{4} [/tex]

[tex] {2}^{4} = {n}^{4} [/tex]

[tex]n = 2[/tex]

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Nomor 3

[tex] = \frac{24 {a}^{ - 7} {b}^{ - 2} c}{6 {a}^{ - 2} {b}^{ - 3} {c}^{ - 6} } [/tex]

[tex] = 4 {a}^{ - 7 - ( - 2)} \: {b}^{ - 2 - ( - 3)} \: {c}^{1 - ( - 6)} [/tex]

[tex] = 4 {a}^{ - 5} \: b \: {c}^{7} [/tex]

[tex] = \frac{4b {c}^{7} }{ {a}^{5} } [/tex]

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Nomor 4

[tex] \frac{ {3}^{2x} }{ {27}^{ - x} } = \sqrt[3]{243} [/tex]

[tex] \frac{ {3}^{2x} }{ {3}^{ - 3x} } = \sqrt[3]{ {3}^{5} } [/tex]

[tex] {3}^{2x - ( - 3x)} = {3}^{ \frac{5}{3} } [/tex]

[tex] {3}^{5x} = {3}^{ \frac{5}{3} } [/tex]

[tex]5x = \frac{5}{3} [/tex]

[tex]x = \frac{ \frac{5}{3} }{5} [/tex]

[tex]x = \frac{5}{3} \times \frac{1}{5} [/tex]

[tex] x = \frac{1}{3} [/tex]