Question

alguien me lo puede resolver xfa xd​

Answer 1

Respuesta: $\quad x=-\frac{7}{8}\quad \left(\mathrm{Decimal}:\quad x=-0.875\right)$

Explicación paso a paso:

$\left(0.375\right)^{x-1}\cdot \left(\frac{1}{3}\right)^{x-1}=\frac{8^{x+2}}{\left(0.25\right)^{x+2}}$

De decimales pasamos a fracciones:

$0.375=\frac{3}{8}$

$0.25=\frac{1}{4}$

Queda:

$\left(\frac{3}{8}\right)^{x-1}\cdot \left(\frac{1}{3}\right)^{x-1}=\frac{8^{x+2}}{\left(\frac{1}{4}\right)^{x+2}}$

$\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \frac{1}{a^b}=a^{-b}$

$\frac{1}{\left(\frac{1}{4}\right)^{x+2}}=\left(\frac{1}{4}\right)^{-x-2}$

$\left(\frac{3}{8}\right)^{x-1}\left(\frac{1}{3}\right)^{x-1}=8^{x+2}\left(\frac{1}{4}\right)^{-x-2}$

$\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \:a^cb^c=\left(ab\right)^c$

$\left(\frac{3}{8}\right)^{x-1}\left(\frac{1}{3}\right)^{x-1}=\left(\frac{3}{8}\cdot \frac{1}{3}\right)^{x-1}=\left(\frac{1}{8}\right)^{x-1}$

$\left(\frac{1}{8}\right)^{x-1}=8^{x+2}\left(\frac{1}{4}\right)^{-x-2}$

$\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \frac{1}{a^b}=a^{-b}$

$\left(\frac{1}{8}\right)^{x-1}=\left(8^{-1}\right)^{x-1}$

$\left(8^{-1}\right)^{x-1}=8^{x+2}\left(\frac{1}{4}\right)^{-x-2}$

$\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \left(a^b\right)^c=a^{bc}$

$\left(8^{-1}\right)^{x-1}=8^{-1\cdot \left(x-1\right)}$

$8^{-1\cdot \left(x-1\right)}=8^{x+2}\left(\frac{1}{4}\right)^{-x-2}$

$\mathrm{Convertir\:}8\mathrm{\:a\:base\:}2$

$8=2^3$

$\left(2^3\right)^{-1\cdot \left(x-1\right)}=\left(2^3\right)^{x+2}\left(\frac{1}{4}\right)^{-x-2}$

$\mathrm{Convertir\:}\frac{1}{4}\mathrm{\:a\:base\:}2$

$\frac{1}{4}=2^{-2}$

$\left(2^3\right)^{-1\cdot \left(x-1\right)}=\left(2^3\right)^{x+2}\left(2^{-2}\right)^{-x-2}$

$\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \left(a^b\right)^c=a^{bc}$

$\left(2^3\right)^{-1\cdot \left(x-1\right)}=2^{3\left(-1\cdot \left(x-1\right)\right)}$

$2^{3\left(-1\cdot \left(x-1\right)\right)}=\left(2^3\right)^{x+2}\left(2^{-2}\right)^{-x-2}$

$\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \left(a^b\right)^c=a^{bc}$

$\left(2^3\right)^{x+2}=2^{3\left(x+2\right)}$

$2^{3\left(-1\cdot \left(x-1\right)\right)}=2^{3\left(x+2\right)}\left(2^{-2}\right)^{-x-2}$

$\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \left(a^b\right)^c=a^{bc}$

$\left(2^{-2}\right)^{-x-2}=2^{-2\left(-x-2\right)}$

$2^{3\left(-1\cdot \left(x-1\right)\right)}=2^{3\left(x+2\right)}\cdot \:2^{-2\left(-x-2\right)}$

$\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \:a^b\cdot \:a^c=a^{b+c}$

$2^{3\left(x+2\right)}\cdot \:2^{-2\left(-x-2\right)}=2^{3\left(x+2\right)-2\left(-x-2\right)}$

$2^{3\left(-1\cdot \left(x-1\right)\right)}=2^{3\left(x+2\right)-2\left(-x-2\right)}$

$\mathrm{Si\:}a^{f\left(x\right)}=a^{g\left(x\right)}\mathrm{,\:entonces\:}f\left(x\right)=g\left(x\right)$

$3\left(-1\cdot \left(x-1\right)\right)=3\left(x+2\right)-2\left(-x-2\right)$

$\mathrm{Simplificar}$

$-3\left(x-1\right)=3\left(x+2\right)-2\left(-x-2\right)$

$\mathrm{Desarrollar\:}-3\left(x-1\right):\quad -3x+3$

$\mathrm{Desarrollar\:}3\left(x+2\right)-2\left(-x-2\right):\quad 5x+10$

$-3x+3=5x+10$

$-3x=5x+7$

$-8x=7$

$\mathrm{Simplificar}$

$x=-\frac{7}{8}$