||
Membaca
SIMAK UI 2010

Solusi SIMAK UI 2010 Kode 209 Nomor 9


Jika\ ^{2}log3=a\ dan\ ^{2}log5=b,\ maka\ ^{30}log\big(75\sqrt[3]{10}\big)=....
(A)\,\,\, \frac{1+a+7b}{3+a+b}
(B)\,\,\, \frac{1+3a+7b}{3+a+b}
(C)\,\,\, \frac{1+3a+7b}{3+3a+3b}
(D)\,\,\, \frac{1+7a+3b}{3+a+b}
(E)\,\,\, \frac{1+7a+3b}{3+3a+3b}
Jawaban: C
\begin{array} {lcl} ^{30}log\big(75\sqrt[3]{10}\big)&=&\frac{^{2}log\big(75\sqrt[3]{10}\big)}{^{2}log30}\\\\&=&\frac{^{2}log\big(75\sqrt[3]{10}\big)}{^{2}log30}\\\\&=&\frac{^{2}log75+^{2}log\sqrt[3]{10}}{^{2}log(2.3.5)}\\\\&=&\frac{^{2}log(3.5^2)+^{2}log10^{\frac{1}{3}}}{^{2}log2+^{2}log3+^{2}log5}\\\\&=&\frac{^{2}log3+2.^{2}log5+\frac{1}{3}.^{2}log10}{1+a+b}\\\\&=&\frac{a+2b+\frac{1}{3}.^{2}log(2.5)}{1+a+b}\\\\&=&\frac{a+2b+\frac{1}{3}(^{2}log2+^{2}log5)}{1+a+b}\\\\&=&\frac{a+2b+\frac{1}{3}(1+b)}{1+a+b}\times \frac{3}{3}\\\\&=&\frac{3a+6b+1+b}{3+3a+3b}\\\\&=&\frac{1+3a+7b}{3+3a+3b} \end {array}

About Kalakay

Guru Matematika SMK

Diskusi

Belum ada komentar.

Tinggalkan Balasan

Isikan data di bawah atau klik salah satu ikon untuk log in:

Logo WordPress.com

You are commenting using your WordPress.com account. Logout / Ubah )

Gambar Twitter

You are commenting using your Twitter account. Logout / Ubah )

Foto Facebook

You are commenting using your Facebook account. Logout / Ubah )

Foto Google+

You are commenting using your Google+ account. Logout / Ubah )

Connecting to %s

%d blogger menyukai ini: