jawablah pertanyaan berikut
(5/6)2 kali (2/5)2
Answer (4)
0\ \ \ and\ \ \ x+1>0\\. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ D=(0;+\infty)\\\\log_3 \frac{x}{x+1} =log_33^2\ \ \ \Leftrightarrow\ \ \ \ \frac{x}{x+1}=9\ /\cdot(x+1)\\\\x=9(x+1)\\\\x=9x+9\\\\-8x=9\ \ \ \Leftrightarrow\ \ \ x=- \frac{9}{8} \ \notin\ D\ \ \ \Rightarrow\ \ \ no\ solution\\\\"> 1 ) l o g 3 x − l o g 3 ( x + 1 ) = 2 l o g 3 3 ⇒ D : x > 0 an d x + 1 > 0 . D = ( 0 ; + ∞ ) l o g 3 x + 1 x = l o g 3 3 2 ⇔ x + 1 x = 9 / ⋅ ( x + 1 ) x = 9 ( x + 1 ) x = 9 x + 9 − 8 x = 9 ⇔ x = − 8 9 ∈ / D ⇒ n o so l u t i o n
0\ \ \ and\ \ \ x+1>0\\. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ D=(0;+\infty)\\\\log \frac{3x}{3(x+1)} =log33^2\ \ \ \Leftrightarrow\ \ \ \frac{x}{(x+1)} =1089\ /\cdot(x+1)\\\\x=1089(x+1)\\\\x=1089x+1089\\\\x-1089x=1089\\\\-1088x=1089\ /:(-1088)\\\\x=- \frac{1089}{1088} \ \notin\ D\ \ \ \Rightarrow\ \ \ no\ solution\\\\Ans.\ the\ equation\ has\ no\ solution."> 2 ) l o g 3 x − l o g 3 ( x + 1 ) = 2 l o g 33 ⇒ D : x > 0 an d x + 1 > 0 . D = ( 0 ; + ∞ ) l o g 3 ( x + 1 ) 3 x = l o g 3 3 2 ⇔ ( x + 1 ) x = 1089 / ⋅ ( x + 1 ) x = 1089 ( x + 1 ) x = 1089 x + 1089 x − 1089 x = 1089 − 1088 x = 1089 / : ( − 1088 ) x = − 1088 1089 ∈ / D ⇒ n o so l u t i o n A n s . t h e e q u a t i o n ha s n o so l u t i o n .
0\ \wedge\ x+1 > 0\\\\x > 0\ \wedge\ x > -1\\\\D:x\in\mathbb{R^+}\\\\(*)\ log\frac{3x}{3(x+1)}=log33^2\iff \frac{x}{x+1}=1089\\\\1089(x+1)=x\\\\1089x+1089-x=0\\\\1088x=-1089\ \ \ \ /:1088\\\\x=-\frac{1089}{1088}\notin D\\\\Answer:no\ solution;\ x\in\O."> l o g 3 x − l o g 3 ( x + 1 ) = 2 l o g 33 ( ∗ ) D : 3 x > 0 ∧ x + 1 > 0 x > 0 ∧ x > − 1 D : x ∈ R + ( ∗ ) l o g 3 ( x + 1 ) 3 x = l o g 3 3 2 ⟺ x + 1 x = 1089 1089 ( x + 1 ) = x 1089 x + 1089 − x = 0 1088 x = − 1089 / : 1088 x = − 1088 1089 ∈ / D A n s w er : n o so l u t i o n ; x ∈ \O .
To solve the equation lo g 3 x − lo g 3 ( x + 1 ) = 2 lo g 3 3 , we combine the logarithms, simplify, and convert to exponential form. The resulting equation leads to a solution of x = − 8 9 , which is not valid since logarithms require positive arguments. Therefore, the equation has no solution.
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[tex]( \frac{5}{6})² \times (\frac{2}{5})² \\ = \frac{25}{36} \times \frac{4}{25} = \frac{100}{900}= \frac{1}{9} \\ [/tex]Jadi, jawaban akhirnya adalah [tex]\boxed{\frac{1}{9}}[/tex]
