Question

quien me lo resuelve tambien xfa pero que nos sea tan largo el procedimiento como la biblia es mucho texto xfa, gracias​

Answer 1

Respuesta:

no se puede tiene que ser largo

Answer 2

Respuesta:  $x=\frac{23-17\log _2\left(3\right)}{2\left(-1+\log _2\left(3\right)\right)}\quad \left(\mathrm{Decimal}:\quad x=-3.37146\dots \right)$

Explicación paso a paso:

$\frac{\left(1.25\right)^{3x-2}}{\left(1\frac{7}{8}\right)^{3x-2}}=\frac{\left(1.125\right)^{5x-3}}{\left(1\frac{11}{16}\right)^{5x+3}}$

Pasamos de decimales a fracciones mixtas:

$\frac{\left(1\frac{1}{4}\right)^{3x-2}}{\left(1\frac{7}{8}\right)^{3x-2}}=\frac{\left(1\frac{1}{8}\right)^{5x-3}}{\left(1\frac{11}{16}\right)^{5x+3}}$

Pasamos de fracciones mixtas a fracciones impropias :

$\frac{\left(\frac{5}{4}\right)^{3x-2}}{\left(\frac{15}{8}\right)^{3x-2}}=\frac{\left(\frac{9}{8}\right)^{5x-3}}{\left(\frac{27}{16}\right)^{5x+3}}$

Resolvemos cad lado:

$\frac{\left(\frac{5}{4}\right)^{3x-2}}{\left(\frac{15}{8}\right)^{3x-2}}$

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

$\frac{\left(\frac{5}{4}\right)^{3x-2}}{\left(\frac{15}{8}\right)^{3x-2}}=\left(\frac{\frac{5}{4}}{\frac{15}{8}}\right)^{3x-2}$

$=\left(\frac{\frac{5}{4}}{\frac{15}{8}}\right)^{3x-2}$

$\frac{\frac{5}{4}}{\frac{15}{8}}=\frac{2}{3}$

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

$\frac{2^{^{3x-2}}}{3^{^{3x-2}}}$

Otro lado :

$\frac{\left(\frac{9}{8}\right)^{5x-3}}{\left(\frac{27}{16}\right)^{5x+3}}$

$\left(\frac{27}{16}\right)^{5x+3}=\frac{27^{5x+3}}{16^{5x+3}}$

$=\frac{\left(\frac{9}{8}\right)^{5x-3}}{\frac{27^{5x+3}}{16^{5x+3}}}$

$\left(\frac{9}{8}\right)^{5x-3}=\frac{9^{5x-3}}{8^{5x-3}}$

$=\frac{\frac{9^{5x-3}}{8^{5x-3}}}{\frac{27^{5x+3}}{16^{5x+3}}}$

$\mathrm{Dividir\:fracciones}:\quad \frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a\cdot \:d}{b\cdot \:c}$

$=\frac{9^{5x-3}\cdot \:16^{5x+3}}{8^{5x-3}\cdot \:27^{5x+3}}$

$\mathrm{Factorizar}\:9^{5x-3}:\quad 3^{2\left(5x-3\right)}$

$\mathrm{Factorizar}\:27^{5x+3}:\quad 3^{3\left(5x+3\right)}$

$\mathrm{Factorizar}\:16^{5x+3}:\quad 2^{4\left(5x+3\right)}$

$\mathrm{Factorizar}\:8^{5x-3}:\quad 2^{3\left(5x-3\right)}$

$=\frac{3^{2\left(5x-3\right)}\cdot \:2^{4\left(5x+3\right)}}{2^{3\left(5x-3\right)}\cdot \:3^{3\left(5x+3\right)}}$

$\mathrm{Cancelar\:}\frac{3^{2\left(5x-3\right)}\cdot \:2^{4\left(5x+3\right)}}{2^{3\left(5x-3\right)}\cdot \:3^{3\left(5x+3\right)}}:\quad 3^{-5x-15}\cdot \:2^{5x+21}$

$=3^{-5x-15}\cdot \:2^{5x+21}$

Igualamos los lados simplificados.

$\frac{2^{^{3x-2}}}{3^{^{3x-2}}}=3^{-5x-15}\cdot \:\:2^{5x+21}$

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

$\frac{1}{3^{3x-2}}=3^{-3x+2}$

$2^{3x-2}\cdot \:3^{-3x+2}=3^{-5x-15}\cdot \:2^{5x+21}$

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

$3^{-3x+2}=\left(2^{\log _2\left(3\right)}\right)^{-3x+2},\:\space3^{-5x-15}=\left(2^{\log _2\left(3\right)}\right)^{-5x-15}$

$2^{3x-2}\left(2^{\log _2\left(3\right)}\right)^{-3x+2}=\left(2^{\log _2\left(3\right)}\right)^{-5x-15}\cdot \:2^{5x+21}$

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

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

$2^{3x-2}\cdot \:2^{\log _2\left(3\right)\left(-3x+2\right)}=\left(2^{\log _2\left(3\right)}\right)^{-5x-15}\cdot \:2^{5x+21}$

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

$\left(2^{\log _2\left(3\right)}\right)^{-5x-15}=2^{\log _2\left(3\right)\left(-5x-15\right)}$

$2^{3x-2}\cdot \:2^{\log _2\left(3\right)\left(-3x+2\right)}=2^{\log _2\left(3\right)\left(-5x-15\right)}\cdot \:2^{5x+21}$

$\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \:a^b\cdot \:a^c=a^{b+c}$$2^{3x-2}\cdot \:2^{\log _2\left(3\right)\left(-3x+2\right)}=2^{3x-2+\log _2\left(3\right)\left(-3x+2\right)},\:\space2^{\log _2\left(3\right)\left(-5x-15\right)}\cdot \:2^{5x+21}=2^{\log _2\left(3\right)\left(-5x-15\right)+5x+21}$$2^{3x-2+\log _2\left(3\right)\left(-3x+2\right)}=2^{\log _2\left(3\right)\left(-5x-15\right)+5x+21}$

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

$3x-2+\log _2\left(3\right)\left(-3x+2\right)=\log _2\left(3\right)\left(-5x-15\right)+5x+21$

$\mathrm{Resolver\:}\:3x-2+\log _2\left(3\right)\left(-3x+2\right)=\log _2\left(3\right)\left(-5x-15\right)+5x+21:\quad x=\frac{23-17\log _2\left(3\right)}{2\left(-1+\log _2\left(3\right)\right)}$

$x=\frac{23-17\log _2\left(3\right)}{2\left(-1+\log _2\left(3\right)\right)}$