PR Matematika buat Besok, Tolong bantuannya. 1. jika log 3 basis 2 = a maka nilai log 9 basis 6 adalah

[tex]^2\log{3}=a$ dan $^3\log{2}=\frac{1}{a} \\ \displaystyle ^6\log{9}=^6\log{3^2}=2\times ^6\log{3} \\ 2\times \frac{1}{^3\log{6}}=2\times \frac{1}{^3\log{3}+^3\log{2}}=2\times\frac{1}{1+\frac{1}{a}} = \frac{2}{\frac{a+1}{a}} \\ \frac{2a}{a+1}[/tex]

²log 3 =a
nilai log 9 basis 6 = 1/^3² log 6
= 2/^3log2 + ^3 log 3
=2/(1/a+1)=2a/1+a