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Author: Ilham D. Matalauni Genre:

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2. Jika $^2{\log }3 = a$    dan $^3{\log }5 = b,$ $^1^5{\log }20 = ...$

a.      $\frac{2}{a}$

b.    $\frac{2 + ab}{a(1+b)}$

c.     $\frac{a}{2}$

d.    $\frac{b + 1}{2ab + 1}$

e.    $\frac{a(1 + b)}{2 + ab}$

Jawaban :

Ingat sifat logaritma :

$^a{\log b } . ^b{\log c } = ^a{\log c }$

$^m{\log a x b } = ^m{\log a } + ^m{\log b }$

$^a{\log b } = \frac{{\log b }}{{\log a }} = \frac{1}{^b{\log a }}$

$^a{\log b } = \frac{1}{n}^a{\log b }$

=>

$^2{\log 3} = a ......... (i)$
$^3{\log 5} = b ......... (ii)$

=>

Dari pers. (i) dan (ii) diperoleh:

$^2{\log 3} . ^3{\log 5} = ^2{\log 5}$

$a . b = ^2{\log 5}$

$^2{\log 5} = a . b ......... (iii)$

$^1^5{\log 20} = ^1^5{\log (4 x 5)}$

$^1^5{\log 4} + ^1^5{\log 5}$

$\frac{1}{^4{\log 15}} + \frac{1}{^5{\log 15}}$

$\frac{1}{^2^{^2}{\log 15}} + \frac{1}{^5{\log 15}}$

$\frac{1}{\frac{1}{2}^2\log 15} + \frac{1}{^5{\log (3 x 5) }}$

$2 . \frac{1}{^2\log 3 + ^2\log 5} + \frac{1}{^5\log 3 + ^5\log 5} .......... (iv)$

=>

Subtitusi pers. (i), (ii), dan (iii) ke pers. (iv), sehingga diperoleh:

$^1^5\log 20 = 2 . \left ( \frac{1}{a + ab} \right ) + \frac{1}{\frac{1}{b} + 1}$

$= \frac{2}{a + ab} + \frac{b}{1 + b}$

$= \frac{2(1 + b) + b(a +ab)}{(a + ab)(1 + b)}$

$= \frac{2(1 + b) + ab(1 +b)}{(a + ab)(1 + b)}$

$= \frac{(1 + b) . (2 + ab)}{a(1 + b)(1 + b)}$

$= \frac{2 + ab}{a(1 + b)}$

Jawaban : option "b". Demikian soal no 2, Next no. 3