Odpowiedź:
[tex]\boxed{a)~~x=1\dfrac{1}{3} }[/tex]
[tex]\boxed{b)~~x=\dfrac{1}{3} }[/tex]
Szczegółowe wyjaśnienie:
Korzystamy z definicji logarytmów oraz wzorów:
- [tex]log_{a} b=c~~\Longleftrightarrow~~a^{c} =b,~~zal.~~a > 0~~\land~~a\neq 0~~\land~~b > 0[/tex]
- [tex]log_{a} b^{c} =c\cdot log_{a} b[/tex]
- [tex]log_{a} a=1[/tex]
Obliczamy:
[tex]\boxed{a)}\\\\5^{3x-2} =25\\\\5^{3x-2} =5^{2} ~~\Longleftrightarrow~~3x-2=2\\\\3x-2=2~~\mid +2\\\\3x=4~~\mid ~\div 3\\\\x=\dfrac{4}{3} \\\\\boxed{x=1\dfrac{1}{3}}[/tex]
sprawdzamy:
[tex]5^{3x-2} =25 ~~\land~~x=\dfrac{4}{3} \\\\L=5^{3\cdot \frac{4}{3} -2}=5^{4-2} =5^{2} =25\\\\P=25\\\\L=P~~~~cbdu[/tex]
[tex]\boxed{b)}\\\\log_{3} x=-1~~\Leftrightarrow~~x=3^{-1} ~~\Rightarrow~~\boxed{x=\dfrac{1}{3}}[/tex]
sprawdzamy:
[tex]log_{3} x=-1~~\land~~x=\dfrac{1}{3}\\\\L=log_{3} \dfrac{1}{3} =log_{3} 3^{-1} =-1\cdot log_{3} 3=-1\cdot 1 =-1\\\\P=-1\\\\L=P~~~~cbdu[/tex]
