Solve for \( x \).

\[ 2 \log_a(3) + \log_a(5) = \log_a(x) \]

0\ \ and\ \ a \neq 1\ \ \ \ \ \ \ D=(0;+\infty) \\\\log_a(3^2)+log_a(5)=log_a(x)\\\\log_a(9\cdot5)=log_ax\ \ \ \Leftrightarrow\ \ \ x=45\ \ \in\ D\\\\Ans.\ x=45"> 2 l o g a ​ ( 3 ) + l o g a ​ ( 5 ) = l o g a ​ ( x ) a > 0 an d a  = 1 D = ( 0 ; + ∞ ) l o g a ​ ( 3 2 ) + l o g a ​ ( 5 ) = l o g a ​ ( x ) l o g a ​ ( 9 ⋅ 5 ) = l o g a ​ x ⇔ x = 45 ∈ D A n s . x = 45

Answered by kate200468 | 2024-06-10

To solve the equation, we use logarithmic properties to combine terms. By expressing the equation as lo g a ​ ( 9 ⋅ 5 ) = lo g a ​ ( x ) , we find that x = 45 .
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Answered by kate200468 | 2024-10-02