Question

Resolver problemas de matemática

Answer 1

Explicação passo-a-passo:

a)

$\sf 2^x = 1024$

$\sf 2^x = 2^{10}$

$\sf \backslash \!\!\! 2^x = \backslash \!\!\! 2^{10}$

$\red{\sf x = 10}$

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b)

$\sf 2^{x+5} = 64$

$\sf 2^{x+5} = 2^6$

$\sf \backslash \!\!\! 2^{x+5} = \backslash \!\!\! 2^6$

$\sf x + 5 = 6$

$\sf x = 6 - 5$

$\red{\sf x = 1}$

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c)

$\sf 3^{x-4} = 243$

$\sf 3^{x-4} = 3^5$

$\sf \backslash \!\!\! 3^{x-4} = \backslash \!\!\! 3^5$

$\sf x - 4 = 5$

$\sf x = 5 + 4$

$\red{\sf x = 9}$

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d)

$\sf 3^{2x-1} = 3^{x+4}$

$\sf \backslash \!\!\! 3^{2x-1} = \backslash \!\!\! 3^{x+4}$

$\sf 2x - 1 = x + 4$

$\sf 2x - x = 4 + 1$

$\red{\sf x = 5}$

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e)

$\sf 2^x = \sqrt[3]{64}$

=> raiz cúbica de 64 = 4

$\sf 2^x = 4$

$\sf 2^x = 2^2$

$\sf \backslash \!\!\! 2^x = \backslash \!\!\! 2^2$

$\red{\sf x = 2}$

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f)

$\sf 4^x = 2^{3x-4}$

$\sf (2^2)^x = 2^{3x-4}$

$\sf 2^{2x} = 2^{3x-4}$

$\sf \backslash \!\!\! 2^{2x} = \backslash \!\!\! 2^{3x-4}$

$\sf 2x = 3x - 4$

$\sf 2x - 3x = - 4$

$\sf - x = - 4~~~*(-1)$

$\red{\sf x = 4}$

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g)

$\sf 5^{x+2} = 125$

$\sf 5^{x+2} = 5^3$

$\sf \backslash \!\!\! 5^{x+2} = \backslash \!\!\! 5^3$

$\sf x + 2 = 3$

$\sf x = 3 - 2$

$\red{\sf x = 1}$

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h)

$\sf 3^x = \sqrt{27}$

=> converta a raiz quadrada de 27 em uma potência fracionária

$\sf 3^x = 27^{\frac{1}{2}}$

$\sf 3^x = (3^3)^{\frac{1}{2}}$

$\sf 3^x = 3^{\frac{3}{2}}$

$\sf \backslash \!\!\! 3^x = \backslash \!\!\! 3^{\frac{3}{2}}$

$\red{\sf x = \dfrac{3}{2}}$

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i)

$\sf 16^{x+2} = 256^{x-4}$

$\sf 16^{x+2} = (16^2)^{x-4}$

$\sf 16^{x+2} = 16^{2x-8}$

$\sf \backslash \!\!\!\! 16^{x+2} = \backslash \!\!\!\! 16^{2x-8}$

$\sf x + 2 = 2x - 8$

$\sf x - 2x = - 8 - 2$

$\sf - x = - 10~~~*(-1)$

$\red{\sf x = 10}$

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j)

$\sf 4^{2x-1} = 32^{3x-8}$

$\sf (2^2)^{2x-1} = (2^5)^{3x-8}$

$\sf 2^{4x-2} = 2^{15x-40}$

$\sf \backslash \!\!\! 2^{4x-2} = \backslash \!\!\! 2^{15x-40}$

$\sf 4x - 2 = 15x - 40$

$\sf 4x - 15x = - 40 + 2$

$\sf - 11x = - 38~~~*(-1)$

$\sf 11x = 38$

$\red{\sf x = \dfrac{38}{11}}$