Pertanyaan

Diketahui 2 lo g 3 = a dan 3 lo g 5 = b . Tentukan nilai c. 9 lo g 120

Diketahui $^{2} lo g 3 = a$ dan $^{3} lo g 5 = b$ . Tentukan nilai

c. $^{9} lo g 120$

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

9 lo g 120 = 2 a a + ab + 2 .

^{9} lo g 120 = \frac{a + ab + 2}{2 a}

Pembahasan

Ingat sifat logaritma Sehingga 9 lo g 120 = = = = = = 3 l o g 9 3 l o g 120 3 l o g ( 3 × 3 ) 3 l o g ( 2 × 2 × 3 × 5 ) 3 l o g 3 + 3 l o g 3 3 l o g 2 + 3 l o g 2 + 3 l o g 3 + 3 l o g 5 1 + 1 a 1 + a 1 + 1 + b 2 a a + ab + 2 2 a a + ab + 2 Dengan demikian, 9 lo g 120 = 2 a a + ab + 2 .

Ingat sifat logaritma

$table attributes columnalign right center left columnspacing 0px end attributes row cell space to the power of a log space b end cell equals cell fraction numerator space to the power of p log space b over denominator space to the power of p log space a end fraction end cell row blank blank cell space to the power of a log space open parentheses b c close parentheses equals to the power of a log space b plus to the power of a log space c end cell row cell space to the power of a log space b end cell equals cell fraction numerator 1 over denominator space to the power of b log space a end fraction end cell end table$

Sehingga

$^{9} lo g 120 = = = = = = \frac{^{3} l o g 120}{^{3} l o g 9} \frac{^{3} l o g ( 2 \times 2 \times 3 \times 5 )}{^{3} l o g ( 3 \times 3 )} \frac{^{3} l o g 2 + ^{3} l o g 2 + ^{3} l o g 3 + ^{3} l o g 5}{^{3} l o g 3 + ^{3} l o g 3} \frac{\frac{1}{a} + \frac{1}{a} + 1 + b}{1 + 1} \frac{\frac{a + ab + 2}{a}}{2} \frac{a + ab + 2}{2 a}$