[tex]\displaystyle\bf\\Folosim~formulele:\\log_{a^m}(b)=\frac{1}{m}log_a(b)\\\\log_a(b^n)=nlog_a(b)\\\\~~\implies~~log_{a^m}(b^n)=\frac{n}{m}log_{a}(b)\\\\log_a(b)=\frac{1}{log_b(a)}\\\\\\Rezolvare:\\\\Metoda~1~de~rezolvare\\\\log_{16}(4)+log_2(4)=\\\\=log_{4^2}(4)+log_2(2^2)=\\\\=\frac{1}{2}log_{4}(4)+2log_2(2)=\\\\=\frac{1}{2}+2=\boxed{\bf2,\!5}[/tex]
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[tex]\displaystyle\bf\\Metoda~2~de~rezolvare:\\\\log_{16}(4)+log_2(4)=\\\\=\frac{1}{log_{4}(16)}+log_2(4)=\\\\=\frac{1}{log_{4}(4^2)}+log_2(2^2)=\\\\=\frac{1}{2log_{4}(4)}+2log_2(2)=\\\\=\frac{1}{2}+2=\boxed{\bf2,\!5}[/tex]
