Como se realiza este logaritmo2Logx - log(x - 6) = 0?
Como se realiza este logaritmo 2Logx - log(x - 6) = 0.
Como se realiza este logaritmo 2Logx - log(x - 6) = 0.
En resumen
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Hola 2. Logx - 2log(x - 1) = 0 dividimos en 2 2. (logx) / 2 - 2(log(x - 1)) / 2 = 0 / 2 log(x) - log(x - 1) = 0 sumamos log(x - 1) log(x) - log(x - 1) + log(x - 1) = 0 + log(x - 1) Al iliminar terminos iguales con…
Log(x ^ 3) = log(6) + 2 log(x) - log(6) - 2 log(x) + log(x ^ 3) = 0 - log(6) - 2 log(x) + log(x ^ 3) = log(1 / 6) + log(1 / x ^ 2) + log(x ^ 3) = log(x ^ 3 / (6 x ^ 2)) = log(x / 6) = log(x / 6) = 0 x / 6 = 1 x = 6…
Tienes que dividirlo.
Sea la expresión : (Logx)² + Logx - 2 = 0 Hacemos cambio de variable Logx = t t² + t - 2 = 0 - - - >Factorizando por el metodo del aspa t 2 = > 2t X t - 1 = > - t - - - - - - t - - - - > Los factores son : (t + 2)(t -…