Решить уравнение:log2(x-4)+log2(x-19)=log2 250 Вычислите:3^2+log3 10. Created by madammasha12. algebra-ru

Решить уравнение:log2(x-4)+log2(x-19)=log2 250 Вычислите:3^2+log3 10...

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$log_2(x-4)+log_2(x-19)=log_2250 \\ log_2(x-4)(x-19)=log_2250 \\ (x-4)(x-19)=250 \\ x^2-4x-19x+76=250 \\ x^2-23x-174=0 \\ D=1225 \\ x_1=-6 \\ x_2=29 \\ x-4\ \textgreater \ 0 \\ x\ \textgreater \ 4 \\ x-19\ \textgreater \ 0 \\ x\ \textgreater \ 19$

x = -6 не удовлетворяет ОДЗ

Ответ. х = 29

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flsh 1)
log₂(x - 4) + log₂(x - 19) = log₂250
$\left \{ {{log_2((x-4)(x-19))=log_2250} \atop {x-4\ \textgreater \ 0}} \right. \\ \left \{ {{(x-4)(x-19)=250} \atop {x\ \textgreater \ 4}} \right. \\ \left \{ {{x^2-23x+76=250} \atop {x\ \textgreater \ 4}} \right. \\ \left \{ {{x^2-23x-174=0} \atop {x\ \textgreater \ 4}} \right. \\ \left \{ {{x_1=-6; x_2=29} \atop {x\ \textgreater \ 4}} \right. \\ x=29$
2)
$3^{2+log_310}=9*3^{log_310}=9*10=90$

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