Решите уравнение log₈(3-2x)-log₈(4-3x)=log₈ 2

㏒₈((3-2x)/(4-3x))=㏒₈2 (3-2x)/(4-3x)=2 3-2x=2*(4-3x) 3-2x=8-6x 4x=5 x=5/4=1.25

log₈(3-2x)-log₈(4-3x)=log₈2 ОДЗ: [latex] \left \{ {{3-2x\ \textgreater \ 0} \atop {4-3x\ \textgreater \ 0}} \right. , \left \{ {{-2x\ \textgreater \ -3} \atop {-3x\ \textgreater \ -4}} \right. , \left \{ {{x\ \textless \ \frac{3}{2} } \atop {x\ \textless \ \frac{4}{3} }} \right. , \left \{ {{x\ \textless \ 1,5} \atop {x\ \textless \ 1 \frac{1}{3} }} \right. [/latex] => [latex]x\ \textless \ 1 \frac{1}{3} [/latex] [latex] log_{8} \frac{3-2x}{4-3x} = log_{8} 2[/latex] [latex] \frac{3-2x}{4-3x} =2[/latex] 3-2x=2*(4-3x) -2x+6x=8-3 4x=5 x=1,25 [latex]1,25 \ \textless \ 1 \frac{1}{3} [/latex] => x=1,25 - корень уравнения