Question

soal Exponen,Tolong di jawab beserta caranya ya, Terimakasih​

Answer 1

a

[tex] {4}^{2x + 3} = \frac{1}{64}[/tex]

[tex] {4}^{2x + 3} = \frac{1}{ {4}^{3} } [/tex]

[tex] {4}^{2x + 3} = {4}^{ - 3} [/tex]

[tex]2x + 3 = - 3[/tex]

[tex]2x + 3 - 3 = - 3- 3[/tex]

[tex]2x = - 6[/tex]

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

[tex]x = - 3[/tex]

b

[tex] {25}^{2x + 1} = {125}^{x - 2} [/tex]

[tex] {5}^{2(2x + 1)} = {5}^{3(x - 2)} [/tex]

[tex] {5}^{4x + 2} = {5}^{3x - 6} [/tex]

[tex]4x + 2 = 3x - 6[/tex]

[tex]4x + 2 - 2 = 3x - 6 - 2[/tex]

[tex]4x = 3x- 8[/tex]

[tex]4x - 3x = 3x - 8 - 3x[/tex]

[tex]x = - 8[/tex]

c

[tex] \frac{ {3}^{x + 1} + {3}^{x + 2} + {3}^{x + 3} }{39} = 27[/tex]

[tex] \frac{ {3}^{x} · {3}^{1} + {3}^{x} · {3}^{2} + {3}^{x}· {3}^{3} }{39} = 27[/tex]

[tex] \frac{ {3}^{x}·3 + {3}^{x} · {3}^{2} + {3}^{x} · {3}^{3} }{39} = 27[/tex]

[tex]39· \frac{ {3}^{x} ·3 + {3}^{x} · {3}^{2} + {3}^{x} · {3}^{3} }{39} = 39·27[/tex]

[tex] {3}^{x} ·3 + {3}^{x} · {3}^{2} + {3}^{x} · {3}^{3} = 39·27[/tex]

[tex] {3}^{x} ·3 + {3}^{x} ·9 + {3}^{x} ·27 = 39·27[/tex]

[tex] {3}^{x} (3 + 9 + 27) = 39·27[/tex]

[tex] {3}^{x} ·39 = 39·27[/tex]

[tex] {3}^{x} = 27[/tex]

[tex] {3}^{x} = {3}^{3} [/tex]

[tex]x = 3[/tex]