1. Zadatak

Prvo rješenje.

Uočimo najprije da je $49=7^2$ i $2401=7^4$.

Primjenjujući svojstva logaritama pojednostavnimo zadani brojni izraz.

$$\begin{aligned} \frac{6-\log_{14}2401}{4-\log_{14}49} &=\frac{6-4\log_{14}7}{4-2\log_{14}7} =\frac{3-2\log_{14}7}{2-\log_{14}7}\\ &=\frac{\log_{14}14^3-\log_{14}7^2}{\log_{14}14^2-\log_{14}7}\\ &=\frac{\log_{14}\frac{14^3}{7^2}}{\log_{14}\frac{14^2}{7}}\\ &=\frac{\log_{14}56}{\log_{14}28}\\ &=\log_{28}56. \end{aligned}$$

Drugo rješenje.

Uočimo najprije da je $49=7^2$ i $2401=7^4$.

Primjenjujući svojstva logaritama pojednostavnimo zadani brojni izraz.

$$\begin{aligned} \frac{6-\log_{14}2401}{4-\log_{14}49} &=\frac{6-4\log_{14}7}{4-2\log_{14}7} =\frac{3-2\log_{14}7}{2-\log_{14}7}\\ &=\frac{3-\frac{2}{\log_7 14}}{2-\frac{1}{\log_7 14}} =\frac{3-\frac{2}{1+\log_7 2}}{2-\frac{1}{1+\log_7 2}}\\ &=\frac{\frac{3+3\log_7 2-2}{1+\log_7 2}}{\frac{2+2\log_7 2-1}{1+\log_7 2}} =\frac{1+3\log_7 2}{1+2\log_7 2}\\ &=\frac{\log_7(7\cdot2^3)}{\log_7(7\cdot2^2)} =\frac{\log_7 56}{\log_7 28}\\ &=\log_{28}56. \end{aligned}$$

Treće rješenje.

Uočimo najprije da je $49=7^2$ i $2401=7^4$.

Primjenjujući svojstva logaritama pojednostavnimo zadani brojni izraz.

$$\begin{aligned} \frac{6-\log_{14}2401}{4-\log_{14}49} &=\frac{6-4\log_{14}7}{4-2\log_{14}7} =\frac{3-2\log_{14}7}{2-\log_{14}7}\\ &=\frac{3-\frac{2\log_{28}7}{\log_{28}14}}{2-\frac{\log_{28}7}{\log_{28}14}}\\ &=\frac{\frac{3\log_{28}14-2\log_{28}7}{\log_{28}14}}{\frac{2\log_{28}14-\log_{28}7}{\log_{28}14}}\\ &=\frac{3\log_{28}14-2\log_{28}7}{2\log_{28}14-\log_{28}7}\\ &=\frac{\log_{28}14^3-\log_{28}7^2}{\log_{28}14^2-\log_{28}7}\\ &=\frac{\log_{28}(14^3:7^2)}{\log_{28}(14^2:7)} =\frac{\log_{28}56}{\log_{28}28}\\ &=\log_{28}56. \end{aligned}$$