Question

Logarytmy - 2 proste zadania proszę o pomoc :)

Answer 1

Oznaczenie: log(a,b) - logarytm o podstawie a z liczby logarytmowanej b.

Zad. 144
Skorzystam ze wzorów:
log(a, b^x) = x * log(a,b)
log(a^y, b) = 1/y * log(a,b).
a) log(2,25) + log(2,25) + log(2,5) = 3log(2,5) + 2log(2,5) + log(2,5) = 6log(2,5)
b) log(3√3,27) = log(3 * 3^(1/2), 3^3) = log(3^(3/2), 3^3) =
   3 * 2/3 log(3,3) = 2*1 = 2
c) log(4√2,128√2) = log(2^2 * 2^(1/2), 2^7 * 2^(1/2)) = log(2^(5/2), 2^(15/2)) = 15/2 * 2/5 * log(2,2) = 3*1 = 3

Zad. 145
a) 5
b) 4
c) 3
d) 3^(-2) = 1/9
e) log(3,2) = -log(1/3, 2); (1/3)^(-log(3,2)) = (1/3)^(log(1/3, 2) = 2
f) 12^(-log(12,4)) = (12^log(12,4))^(-1) = 4^(-1) = 1/4
g) 6^(-log(6,3))^(-5) = (6^log(6,3))^5 = 3^5
h) 10^log(10,4) = 4; 4^log(4,11) = 11.