Kamis, 09 September 2010

Pembahasan Logaritma

Berikut contoh pembahasan soal logartima yang diambil dari soal Ujian Nasional.



1. Nilai dari \large \frac{^{3}\log \sqrt{6}}{(^{3}\log 18)^{2}-(^{3}\log 2)^{2}} = …. (UN 2010)

Pembahasan:
\large \frac{^{3}\log \sqrt{6}}{(^{3}\log 18)^{2}-(^{3}\log 2)^{2}}
= \large \frac{^{3}\log 6^\frac{1}{2}}{(^{3}\log 18+^{3}\log 2)(^{3}\log 18 - ^{3}\log 2)}, ingat a^{2}-b^{2}=(a+b)(a-b)
=\frac{1}{2}\large \frac{^{3}\log 6}{^{3}\log36.^{3}\log9}
= \frac{1}{2}\large \frac{^{3}\log 6}{^{3}\log(6)^{2}.2}
= \frac{1}{2}\large \frac{^{3}\log 6}{2.^{3}\log 6.2}
= \frac{1}{2}\large \frac{1}{2.2}= \frac{1}{8}
Jadi nilai dari \frac{^{3}\log \sqrt{6}}{(^{3}\log 18)^{2}-(^{3}\log 2)^{2}} =\frac{1}{8}


2. Nilai dari \large \frac{^{27}\log 9 + ^{2} \log 3.^{\sqrt{3}}\log 4}{^{3} \log 2 - ^{3} \log 18} = ....  (UN 2010).

Pembahasan:
\large \frac{^{27}\log 9 + ^{2} \log 3.^{\sqrt{3}}\log 4}{^{3} \log 2 - ^{3} \log 18}
= \large \frac{^{(3)^{3}}\log (3)^{2} + ^{2} \log 3.^{(3)^{\frac{1}{2}}}\log 4}{^{3} \log 2 - ^{3} \log 18}
= \large \frac{\frac{2}{3}.^{3}\log 3 + ^{2} \log 3.^{(3)^{\frac{1}{2}}}\log (2)^{2}}{^{3} \log(\frac{2}{18})}
= \large \frac{\frac{2}{3}.^{3}\log 3 + ^{2} \log 3.\frac{2}{\frac{1}{2}}.^{3}\log (2)}{^{3} \log(\frac{2}{18})}
= \large \frac{\frac{2}{3} + 4 .^{2} \log 3.^{3}\log 2}{^{3} \log(\frac{1}{9})}
= \large \frac{\frac{2}{3} + 4 }{-2}
= \large \frac{\frac{14}{3}}{-2}= -\frac{14}{6}
Jadi nilai dari \large \frac{^{27}\log 9 + ^{2} \log 3.^{\sqrt{3}}\log 4}{^{3} \log 2 - ^{3} \log 18} = -\frac{14}{6}.


3. Untuk x yang memenuhi \large ^{2}\log 16^{\frac{2x-1}{4}}= 8, maka 32 x =  ….   (UN 2009)

Pembahasan:
\large ^{2}\log 16^{\frac{2x-1}{4}}= 8, \Leftrightarrow \large 2^{8} = 16^{\frac{2x-1}{4}}
\large 2^{8} = ((2)^{4})^{\frac{2x-1}{4}}
\large 2^{8} = (2)^{2x-1}, sehingga
2x - 1 = 8
x = \frac {9}{2}
maka nilai 32x = 32.\frac{9}{2}= 144.


4. Hasil dari \large ^{\frac{1}{5}}\log 625 + ^{64}\log \frac{1}{16}+ 4^{3^{^{25}\log 5}}=  …. (UN 2010)
Pembahasan
\large ^{\frac{1}{5}}\log 625 + ^{64}\log \frac{1}{16}+ 4^{3^{^{25}\log 5}}
= \large ^{(5)^{-1}}\log (5)^{4} + ^{(4)^{3}}\log (4)^{-2}+ 4^{(3)^{\frac{1}{2}}}
=\large \frac{4}{-1}.^{5}\log 5 + \frac{-2}{3}.^{4}\log 4 + 4^{(\frac{3}{2})}
=\large -4- \frac{2}{3}+ (2)^{3}
=\large -4- \frac{2}{3}+ 8 = 3\frac{1}{3}.
Jadi hasil dari \large ^{\frac{1}{5}}\log 625 + ^{64}\log \frac{1}{16}+ 4^{3^{^{25}\log 5}}= 3\frac{1}{3}.

Contoh pengerjaan sederhana logaritma dapat dilihat di sini.
(format power point .ppt)

warna coretan ,