Jika 2 log 3 = a dan 3 log 5 =b nyatakan bentuk berikut dalam a dan b

a. 30 log 150
b. 100 log 50

Jika 2 log 3 = a dan 3 log 5 =b nyatakan bentuk berikut dalam a dan b

Jawaban Terkonfirmasi

A. 30 log 150
3 log 150
3 log 30

3 log 5.5.2.3
3 log 5.2.3

3 log 5 + 3 log 5 + 3 log 2 + 3 log 3
3 log 5 + 3 log 2 + 3 log 3

b + b + 1/a + 1 × a
b + 1/a + 1 a

ab + ab + 1 + a
ab + 1 + a

⇒ 2ab + 1 + a
ab + 1 + a

b. 100 log 50
3 log 50
3 log 100

3 log 5.5.2
3 log 5.5.2.2

3 log 5 + 3 log 5 + 3 log 2
3 log 5 + 3 log 5 + 3 log 2 + 3 log 2

b + b + 1/a × a
b + b + 1/a + 1/a a

ab + ab + 1
ab + ab + 1 + 1

⇒ 2ab + 1
2ab + 2

maaf kalo salah 🙂

Jawaban Terkonfirmasi

$bold{underline{Solution a :}}\\~^{30}log 150\=~^{30}log 3 times ~^{3}log 150\= frac{1}{~^{3}log 30} times ~^{3}log 150\= frac{~^{3}log 150}{~^{3}log 30}\= frac{~^{3}log (3 times 25 times 2)}{~^{3}log (3 times 2 times 5)}\= frac{~^{3}log 3~+ ~^{3}log 25~+~^{3}log 2}{~^{3}log 3 ~+~^{3}log 2~+~^{3}log 5}\= frac{~^{3}log 3~+ ~^{3}log 5^{2}~+~^{3}log 2}{~^{3}log 3 ~+~^{3}log 2~+~^{3}log 5}$
$\= frac{~^{3}log 3~+ 2~^{3}log 5~+~^{3}log 2}{~^{3}log 3 ~+~^{3}log 2~+~^{3}log 5}\=frac{1+2(b)+ frac{1}{a}}{1+ frac{1}{a}+b}\=frac{1+2b+ frac{1}{a}}{1+ frac{1}{a}+b} times frac{a}{a}\=frac{a+2ab+ 1}{a+ 1+ab}$

$bold{underline{solution b :}}\\~^{100}log 50\=~^{100}log 3 times ~^{3}log 50\= frac{1}{~^{3}log 100} times ~^{3}log 50\= frac{~^{3}log 50}{~^{3}log 100}\= frac{~^{3}log (2 times 25)}{~^{3}log (4 times 25)}\= frac{~^{3}log 2~+ ~^{3}log 25}{~^{3}log 4~+~^{3}log 25}\= frac{~^{3}log 2~+ ~^{3}log 5^{2}}{~^{3}log 2^{2}~+~^{3}log 5^{2}}\= frac{~^{3}log 2~+ 2~^{3}log 5}{2~^{3}log 2~+2~^{3}log 5}\= frac{ frac{1}{a} ~+ 2(b)}{2~( frac{1}{a} )~+2(b)}$
$\= frac{ frac{1}{a} ~+ 2b}{frac{2}{a}~+2b} times frac{a}{a}\= frac{ 1~+ 2ab}{2~+2ab}$

$======================================\underline{bold{Note:}}\\remember this formula:\1]boxed{~^{a}log b times ~^{b}log c=~^{a}log c}\\2]boxed{~^{a}log (b times c)=~^{a}log b+~^{a}log c}\\3]boxed{~^{a}log b= frac{1}{~^{b}log a} }\\4]boxed{~^{a}log a = 1}\\5]boxed{~^{a}log b^{m}=m~^{a}log b}$