Question

ПРОШУ ПОМОГИТЕ МНЕ С ЛОГАРИФМАМИ,УМОЛЯЮЮ 1. log2(a/b),если log2(a-7):log2(b-5) 2. log2(m/4n),если log2(m-3):log2(n-7) 3.log3(k/27p),если log3(k-4):log3(p-7)

Answer 1

1. $log_{2}$(a/b)=$log_{2}$a-$log_{2}$b=7-5=2

2.$log_{2}$(m/4n)=$log_{2}$m-$log_{2}$4n=$log_{2}$m-($log_{2}$4+$log_{2}$n)=3-(2+7)=3-9=-6


3.$log_{3}$(k/27p)=$log_{3}$k-$log_{3}$27p=$log_{3}$k-($log_{3}$27+$log_{3}$p)=4-(3+7)=4-10=-6

Answer 2

Вопрос 1

log2(a/b) = log2(a) - log2(b) Если log2(a-7) = log2(b-5) + 2, то: log2(a) - log2(b) = log2(b-5) + 2 log2(a) = log2(b) + log2(b-5) + 2 log2(a) = 3 a = 2^3 = 8

Вопрос 2

log2(m/4n) = log2(m) - 2log2(n) Если log2(m-3) = log2(n-7), то: log2(m) - 2log2(n) = log2(n-7) log2(m) = 2log2(n) + log2(n-7) log2(m) = 3 m = 2^3 = 8

Вопрос 3

log3(k/27p) = log3(k) - 2log3(p) Если log3(k-4) = log3(p-7), то: log3(k) - 2log3(p) = log3(p-7) log3(k) = 2log3(p) + log3(p-7) log3(k) = 3 k = 3^3 = 27