Nu vreau raspunsul direct,ci cu explicații!Dau coroana!
[tex] log_3{5} + log_3{6} - log_3{6}[/tex]
[tex] log_3{3} + log_6{36}[/tex]
[tex] log_3{27} - log_2{32}[/tex]
[tex] log_4{64}[/tex]
[tex] log_3{3}[/tex]

Răspuns :

Sa rezolvam:

a)

[tex]\log_35+\log_36-\log_36= \text{\((\log_36\) se reduce cu \(-\log_36\)) }\log_35[/tex]

b)

[tex]\log_33=\log_3\left(3^1\right)=1\\\log_636=\log_6\left(6^2\right)=2\\\text{Si in general: }log_a\left(a^n\right)=n\\\log_33+\log_636=1+2=3[/tex]

c)

[tex]\log_327=\log_3\left(3^3\right)=3\\\log_232=\log_2\left(2^5\right)=5\\\log_327-\log_232=3-5=-2[/tex]

d)

[tex]\log_464=\log_4\left(4^3\right)=3[/tex]

e)

[tex]\log_33=\log_3\left(3^1\right)=1[/tex]