5 log50+2log64+5log30-5log12

#semoga membantu -^^-

[tex]Untuk\ diingat: \\^{x}log\ a+^{x}log\ b=^{x}log\ (a \times b) \\^{x}log\ a-^{x}log\ b=^{x}log\ (\frac{a}{b})\\ ^{x}log\ a^b=b \times ^{x}log\ a\\ ^{a}log\ a=1[/tex]

[tex]=\ ^{5}log\ 50+\ ^{2}log\ 64+\ ^{5}log\ 30-\ ^{5}log\ 12\\ \\=(^{5}log\ 50+\ ^{5}log\ 30-^{5}log\ 12)+\ ^{2}log\ 64\\ \\=(^{5}log\ (\frac{50 \times 30}{12}))+\ ^{2}log\ 2^6\\ \\=(^{5}log\ (\frac{1500}{12}))+6\ ^{2}log\ 2\\ \\=\ ^{5}log\ 125+6\\ \\=\ ^{5}log\ (25 \times 5)+6 \\ \\=\ ^{5}log\ 25+\ ^{5}log\ 5+6[/tex]

[tex]=\ ^{5}log\ 5^2+1+6 \\ \\=(2 \times ^{5}log\ 5)+7 \\ \\ =(2 \times 1)+7 \\ =2+7 \\ =9 \\ \\ \\ Jadi: \\ ^{5}log\ 50+\ ^{2}log\ 64+\ ^{5}log\ 30-\ ^{5}log\ 12=9 [/tex]