log2(15+x)=log2^3 la ecuación
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Solución
Solución detallada
Tenemos la ecuación
log ( x + 15 ) log ( 2 ) = log ( 2 ) 3 \frac{\log{\left(x + 15 \right)}}{\log{\left(2 \right)}} = \log{\left(2 \right)}^{3} log ( 2 ) log ( x + 15 ) = log ( 2 ) 3 log ( x + 15 ) log ( 2 ) = log ( 2 ) 3 \frac{\log{\left(x + 15 \right)}}{\log{\left(2 \right)}} = \log{\left(2 \right)}^{3} log ( 2 ) log ( x + 15 ) = log ( 2 ) 3 Devidimos ambás partes de la ecuación por el multiplicador de log =1/log(2)
log ( x + 15 ) = log ( 2 ) 4 \log{\left(x + 15 \right)} = \log{\left(2 \right)}^{4} log ( x + 15 ) = log ( 2 ) 4 Es la ecuación de la forma:
log(v)=p Por definición log
v=e^p entonces
x + 15 = e log ( 2 ) 3 1 log ( 2 ) x + 15 = e^{\frac{\log{\left(2 \right)}^{3}}{\frac{1}{\log{\left(2 \right)}}}} x + 15 = e l o g ( 2 ) 1 l o g ( 2 ) 3 simplificamos
x + 15 = e log ( 2 ) 4 x + 15 = e^{\log{\left(2 \right)}^{4}} x + 15 = e l o g ( 2 ) 4 x = − 15 + e log ( 2 ) 4 x = -15 + e^{\log{\left(2 \right)}^{4}} x = − 15 + e l o g ( 2 ) 4
Gráfica
-35.0 -32.5 -30.0 -27.5 -25.0 -22.5 -20.0 -17.5 -15.0 -12.5 -10.0 -7.5 -5.0 -20 20
Suma y producto de raíces
[src]
− 15 + e log ( 2 ) 4 -15 + e^{\log{\left(2 \right)}^{4}} − 15 + e l o g ( 2 ) 4
− 15 + e log ( 2 ) 4 -15 + e^{\log{\left(2 \right)}^{4}} − 15 + e l o g ( 2 ) 4
− 15 + e log ( 2 ) 4 -15 + e^{\log{\left(2 \right)}^{4}} − 15 + e l o g ( 2 ) 4
− 15 + e log ( 2 ) 4 -15 + e^{\log{\left(2 \right)}^{4}} − 15 + e l o g ( 2 ) 4
x 1 = − 15 + e log ( 2 ) 4 x_{1} = -15 + e^{\log{\left(2 \right)}^{4}} x 1 = − 15 + e l o g ( 2 ) 4