3)Sa se arate că:
log(2)14+log(2)3-log(2)6=log(2)7
log(2)14•3:6=log(2)7
log(2)42:6=log(2)7
log(2)7=log(2)7
4)Sa se compare numerele:
[tex]( \frac{1}{4} ) {}^{ - 2} = {4}^{2} = 16 \\ \sqrt{64} = \sqrt{ {8}^{2} } = 8 \\ \sqrt[3]{8} = \sqrt[3]{ {2}^{3} } =2 \\ 16 > 8 > 2 \\( \frac{1}{4} ) {}^{ - 2} > \sqrt{64} > \sqrt[3]{2} [/tex]
