Riješite zadatke.

Rješavamo logaritamsku jednadžbu:
$\log(2-d) + 1 = 3$$\log(2-d) + 1 = 3$
$\log(2-d) = 2$$\log(2-d) = 2$
Definicija logaritma ($10^2 = ...$$10^2 = ...$):
$2 - d = 100$$2 - d = 100$
$-d = 98 \implies d = -98$$-d = 98 \implies d = -98$

Zadano: $\log_a 2 + \log_a b = 1$$\log_a 2 + \log_a b = 1$ i $\log_b a = 2$$\log_b a = 2$.
Iz druge jednakosti: $a = b^2$$a = b^2$ ili $b = a^{1/2}$$b = a^{1/2}$.
Uvrstimo u prvu jednadžbu (sve svedemo na bazu $a$$a$):
$\log_a 2 + \log_a (a^{1/2}) = 1$$\log_a 2 + \log_a (a^{1/2}) = 1$
$\log_a 2 + \frac{1}{2} = 1$$\log_a 2 + \frac{1}{2} = 1$
$\log_a 2 = \frac{1}{2} \implies a^{1/2} = 2 \implies a = 4$$\log_a 2 = \frac{1}{2} \implies a^{1/2} = 2 \implies a = 4$.