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2log2(x + 1) - log4(x + 1) = 1?

2log2(x + 1) - log4(x + 1) = 1.

Pregunta de Matemáticas · 1 respuesta · 4 votos · mejor respuesta de Ingrixitha

En resumen

2log2(x + 1) - log4(x + 1) = 1 2 * [log2 + log(x + 1)] - [log4 + log(x + 1)] = 1 2log2 + 2log(x + 1) - log4 - log(x + 1) = 1 2log2 + 2log(x + 1) - log2² - log(x + 1) = 1 2log2 + 2log(x + 1) - 2log2 - log(x + 1) = 1 2log(x + 1) - log(x + 1) = 1 log(x + 1) = 1 10 = x + 1 x = 9.

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Mejor respuesta

Ingrixitha
10

2log2(x + 1) - log4(x + 1) = 1

2 * [log2 + log(x + 1)] - [log4 + log(x + 1)] = 1

2log2 + 2log(x + 1) - log4 - log(x + 1) = 1

2log2 + 2log(x + 1) - log2² - log(x + 1) = 1

2log2 + 2log(x + 1) - 2log2 - log(x + 1) = 1

2log(x + 1) - log(x + 1) = 1

log(x + 1) = 1

10 = x + 1

x = 9.