Помогите, пожалуйста, с 16, 18,19,20 и 21 заданием. Просто не доходит до меня

16) 25^[2 \ log(V3) 5] = =>
___ log(V3) 5 = 1 \ log(5) V3 = 1 \ log(5) 3^(1\2) = 1 \ (1\2)*log(5) 3 = 2 \ log(5) 3
___ 2 \ log(V3) 5 = 2 \ [2 \ log(5) 3] = 2 * log(5) 3 \ 2 = log(5) 3
___ 25 = 5^2
=>
= (5^2)^[log(5) 3] =
= 5^[2 * log(5) 3] =
= 5^[log(5) 3^2] = 3^2 = 9

18) (V5)^(2+lg 196) * 2^(lg 2 + lg 7) =
= [5^(1\2)]^(2+lg 196) * 2^(lg 2*7) =
= 5^[1\2 * (2+lg 196)] * 2^(lg 14)
= 5^(1 + 1\2 * lg 196) * 2^(lg 14) =
= 5^[1 + lg 196^(1\2)] * 2^(lg 14) =
= 5 * 5^(lg 14) * 2^(lg 14) =
= 5 * (5*2)^(lg 14) =
= 5 * 10^(lg 14) =
= 5 * 14 = 70

19) log(5) 2 = a; log(5) 3 = b -----> log(3) 4 = ?
log(5) 2 = a ------> 5^a = 2 ------> (5^a)^2 = 2^2 ------> 4 = 5^(2a)
log(5) 3 = b ------> 3 = 5^(b)
=>
log(5^b) {5^(2a)} =
= 1\b * 2a * log(5) 5 = 2a\b * 1 = 2a\b

21) log(a) b = 3 ------> log(ab) b = ?
log(a) b = 3 -------> a^3 = b
=>
log(ab) b = 1 \ log(b) {ab} = 1 \ [log(b) a + log(b) b] = 1 \ [log(b) a + 1]
=>
log(ab) b = 1 \ [log(a^3) a + 1] = 1 \ [1\3 * log(a) a +1] = 1 \ [1\3 + 1] = 1 \ (4\3) = 3\4

21) log(4) 5 * log(5) 6 * log(6) 7 * log(7) 32 = =>
___ log(4) 5 * log(5) 6 = 1\log(5) 4 * log(5) 6 = log(5) 6 \ log(5) 4 = log(4) 6
___ log(6) 7 * log(7) 32 = 1\log(7) 6 * log(7) 32 = log(7) 32 \ log(7) 6 = log(6) 32
=>
= log(4) 6 * log(6) 32 =
= 1\log(6) 4 * log(6) 32 =
= log(6) 32 \ log(6) 4 =
= log(4) 32 =
= log(2^2) 2^5 =
= 1\2 * 5 * log(2) 2 = 5\2 * 1 = 5\2

до нас тоже не доходит. Где мы обязаны отыскать это задание?