Question

Resolver la ecuación

Answer 1

La ecuación da como resultado: $x=0.84842$

Resolución por pasos

$2^{3x+1}\cdot \:5^{4x-2}=110$

Ley de los exponentes: $a=b^{\log _b\left(a\right)}$

$2^{3x+1}\left(2^{\log _2\left(5\right)}\right)^{4x-2}=110$

Ley de los Exponentes: $\left(a^b\right)^c=a^{bc}$

$2^{3x+1}\cdot \:2^{\log _2\left(5\right)\left(4x-2\right)}=110$

Ley de los Exponentes :$a^b\cdot \:a^c=a^{b+c}$

SI $f\left(x\right)=g\left(x\right)$ entonces $\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)$

$\ln \left(2^{3x+1+\log _2\left(5\right)\left(4x-2\right)}\right)=\ln \left(110\right)$

Aplicar propiedad de logaritmo:$\log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$\left(3x+1+\log _2\left(5\right)\left(4x-2\right)\right)\ln \left(2\right)=\ln \left(110\right)$

Dividir entre $\ln \left(2\right)$

$3x+1+\log _2\left(5\right)\left(4x-2\right)=\frac{\ln \left(110\right)}{\ln \left(2\right)}$

$3x+\log _2\left(5\right)\left(4x-2\right)=\frac{\ln \left(110\right)}{\ln \left(2\right)}-1$

$3x+4\log _2\left(5\right)x-2\log _2\left(5\right)=\frac{\ln \left(110\right)}{\ln \left(2\right)}-1$

Siempre tratando de despejar x

$3x+4\log _2\left(5\right)x=\frac{\ln \left(110\right)}{\ln \left(2\right)}-1+2\log _2\left(5\right)$

Factor comun x

$\left(3+4\log _2\left(5\right)\right)x=\frac{\ln \left(110\right)}{\ln \left(2\right)}-1+2\log _2\left(5\right)$

Dividir ambos lados entre: $3+4\log _2\left(5\right)$

$\frac{\left(3+4\log _2\left(5\right)\right)x}{3+4\log _2\left(5\right)}=\frac{\frac{\ln \left(110\right)}{\ln \left(2\right)}}{3+4\log _2\left(5\right)}-\frac{1}{3+4\log _2\left(5\right)}+\frac{2\log _2\left(5\right)}{3+4\log _2\left(5\right)}$

Simplifico $\frac{\left(3+4\log _2\left(5\right)\right)x}{3+4\log _2\left(5\right)}:\quad x$

Simplificando: $\frac{\frac{\ln \left(110\right)}{\ln \left(2\right)}}{3+4\log _2\left(5\right)}-\frac{1}{3+4\log _2\left(5\right)}+\frac{2\log _2\left(5\right)}{3+4\log _2\left(5\right)}$

$=\frac{\frac{\ln \left(110\right)}{\ln \left(2\right)}-1+2\log _2\left(5\right)}{3+4\log _2\left(5\right)}$

$=\frac{\frac{\ln \left(55\right)+2\ln \left(5\right)}{\ln \left(2\right)}}{3+4\log _2\left(5\right)}=\frac{\ln \left(55\right)+2\ln \left(5\right)}{\ln \left(2\right)\left(3+4\log _2\left(5\right)\right)}$

$=\frac{\ln \left(1375\right)}{\ln \left(2\right)\left(4\log _2\left(5\right)+3\right)}=\frac{\ln \left(1375\right)}{\ln \left(5000\right)}$

Entonces:

$x=\frac{\ln \left(1375\right)}{\ln \left(5000\right)}$

$x=0.84842$