[tex]\displaystyle \sf 3^{x-1}=5^{3-x} \\\\ \text{Aplicando log na base "e" em ambos os lados } : \\\\ \ln 3^{x-1} = \ln 5^{3-x} \\\\ (x-1)\cdot \ln3 =(3-x)\cdot \ln 5 \\\\ x\cdot \ln3-\ln 3 = 3\cdot \ln 5-x\cdot \ln 5 \\\\ x\cdot \ln 3 +x\cdot \ln5= \ln 5^3 +\ln 3 \\\\ x\cdot (\ln 3+\ln5) = \ln (125\cdot 3) \\\\ x\cdot \ln (5\cdot 3) = \ln (375) \\\\\\ \huge\boxed{\sf x = \frac{\ln( 375)}{\ln (15) } \ }\checkmark[/tex]
OU

[tex]\displaystyle \sf \huge\boxed{\sf x = \log_{\ 15}\ (375)\ }\checkmark[/tex]

Resposta:

[tex]\textsf{Leia abaixo}[/tex]

Explicação passo a passo:

[tex]\mathsf{3^{x - 1} = 5^{3 - x}}[/tex]

[tex]\mathsf{\dfrac{3^x}{3} = \dfrac{5^3}{5^x}}[/tex]

[tex]\mathsf{(15)^x = 375}[/tex]

[tex]\mathsf{log\:(15)^x = log\:375}[/tex]

[tex]\mathsf{x\:log\:15 = log\:375}[/tex]

[tex]\boxed{\boxed{\mathsf{x = \dfrac{log\:375}{log\:15}}}}[/tex]