Tolong jawabkan ya kak beserta caranya tentang logaritma dassar

Diketahui:

[tex]^{2}\log3=a\qquad\text{dan}\quad ^{3}\log5=b[/tex]

[tex]^{18}\log50=\dfrac{^{3}\log50}{^{3}\log18} \\ \\ ^{3}\log50=^{3}\log\left(2\cdot5^{2}\right) \\ ^{3}\log50=^{3}\log2+^{3}\log5^{2} \\ ^{3}\log50=^{3}\log2+2\left(^{3}\log5\right) \\ ^{3}\log50=^{3}\log2+2b[/tex]

[tex]^{3}\log18=^{3}\log\left(2\cdot3^{2}\right) \\^{3}\log18 =^{3}\log2+^{3}\log3^{2} \\ ^{3}\log18=^{3}\log2+2\left(^{3}\log3\right) \\ ^{3}\log18=^{3}\log2+2\cdot1 \\ ^{3}\log18=^{3}\log2+2[/tex]

Karena, [tex]^{3}\log2=\frac{^{2}\log2}{^{2}\log3}=\frac{1}{a}[/tex] maka:

[tex]^{18}\log50=\dfrac{^{3}\log50}{^{3}\log18} \\ ^{18}\log50=\frac{^{3}\log2+2b}{^{3}\log2+2} \\ ^{18}\log50=\dfrac{\frac{1}{a}+2b}{\frac{1}{a}+2} \\ ^{18}\log50=\dfrac{\frac{1+2ab}{a}}{\frac{1+2a}{a}} \\ \\ ^{18}\log50=\dfrac{1+2ab}{1+2a}[/tex]

Jawabannya yang D ya.

Untuk soal yang satunya:

[tex]^{5}\log\left(5\sqrt{5}\right)=^{5}\log5+^{5}\log\sqrt{5} \\ ^{5}\log\left(5\sqrt{5}\right)=1+^{5}\log5^{\frac{1}{2}} \\ ^{5}\log\left(5\sqrt{5}\right) =1+\frac{1}{2}\left(^{5}\log5\right) \\ ^{5}\log\left(5\sqrt{5}\right)=1+\frac{1}{2}(1) \\ ^{5}\log\left(5\sqrt{5}\right)=1+\frac{1}{2}[/tex]