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ln(v)=e^x la ecuación

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Solución

Ha introducido [src]
          x
log(v) = E 
log(v)=ex\log{\left(v \right)} = e^{x}
Solución detallada
Tenemos la ecuación:
log(v)=ex\log{\left(v \right)} = e^{x}
o
ex+log(v)=0- e^{x} + \log{\left(v \right)} = 0
o
ex=log(v)- e^{x} = - \log{\left(v \right)}
o
ex=log(v)e^{x} = \log{\left(v \right)}
- es la ecuación exponencial más simple
Sustituimos
v=exv = e^{x}
obtendremos
vlog(v)=0v - \log{\left(v \right)} = 0
o
vlog(v)=0v - \log{\left(v \right)} = 0
hacemos cambio inverso
ex=ve^{x} = v
o
x=log(v)x = \log{\left(v \right)}
Entonces la respuesta definitiva es
x1=log(log(v))log(e)=log(log(v))x_{1} = \frac{\log{\left(\log{\left(v \right)} \right)}}{\log{\left(e \right)}} = \log{\left(\log{\left(v \right)} \right)}
Gráfica
Suma y producto de raíces [src]
suma
I*arg(log(v)) + log(|log(v)|)
log(log(v))+iarg(log(v))\log{\left(\left|{\log{\left(v \right)}}\right| \right)} + i \arg{\left(\log{\left(v \right)} \right)}
=
I*arg(log(v)) + log(|log(v)|)
log(log(v))+iarg(log(v))\log{\left(\left|{\log{\left(v \right)}}\right| \right)} + i \arg{\left(\log{\left(v \right)} \right)}
producto
I*arg(log(v)) + log(|log(v)|)
log(log(v))+iarg(log(v))\log{\left(\left|{\log{\left(v \right)}}\right| \right)} + i \arg{\left(\log{\left(v \right)} \right)}
=
I*arg(log(v)) + log(|log(v)|)
log(log(v))+iarg(log(v))\log{\left(\left|{\log{\left(v \right)}}\right| \right)} + i \arg{\left(\log{\left(v \right)} \right)}
i*arg(log(v)) + log(Abs(log(v)))
Respuesta rápida [src]
x1 = I*arg(log(v)) + log(|log(v)|)
x1=log(log(v))+iarg(log(v))x_{1} = \log{\left(\left|{\log{\left(v \right)}}\right| \right)} + i \arg{\left(\log{\left(v \right)} \right)}
x1 = log(Abs(log(v))) + i*arg(log(v))