Diketahui 3log2 = p. Nilai dari 8log12 =

8log12 = 8log4 x 3

= 8log4 + 8log3
= 2³log2² + 2³log3
= [tex] \frac{2}{3}[/tex] + [tex] \frac{1}{3}[/tex] x 2log3
= [tex] \frac{2}{3}[/tex] + [tex] \frac{1}{3}[/tex] x [tex] \frac{1}{p}[/tex]
= [tex] \frac{2}{3}[/tex] + [tex] \frac{1}{3p}[/tex]
= [tex] \frac{2p + 1}{3p}[/tex]

[tex]log_3 {2} = p[/tex]
Tentukan [tex]log_8 {12}[/tex] dalam bentuk p

[tex]log_8 {12} = \frac{log 12}{log 8} \\ log_8 {12} = \frac{log_3 {(2)(2)(3)}}{log_3 {(2)(2)(2)}} \\ log_8 {12} = \frac{log_3 {2} + log_3{2} + log_3 {3}}{log_3{2}+log_3{2}+log_3{2}} \\ log_8 {12} = \frac{p+ p + 1}{p + p + p} \\ log_8 {12} = \frac{2p+1}{3p} \\[/tex]