Solve for x:

\[
\log 6x - \log 2 = 1
\]

0\to x > 0\to x\in\mathbb{R^+}\\\\log6x-log2=1\\\\log\left(6x:2\right)=log10^1\\\\3x=10\ /:3\\\\x=\frac{10}{3}\\\\x=3\frac{1}{3}\in D"> l o g 6 x − l o g 2 = 1 ; D : 6 x > 0 → x > 0 → x ∈ R + l o g 6 x − l o g 2 = 1 l o g ( 6 x : 2 ) = l o g 1 0 1 3 x = 10 / : 3 x = 3 10 ​ x = 3 3 1 ​ ∈ D
0\ \wedge\ b > 0\ \wedge\ c > 0\ \wedge\ d\in\mathbb{R}\\\\log_ab-log_ac=log_a(b:c)\\\\log_ab=d\iff\ a^d=b\ \ (log_ab=log_aa^d)"> a > 0 ∧ b > 0 ∧ c > 0 ∧ d ∈ R l o g a ​ b − l o g a ​ c = l o g a ​ ( b : c ) l o g a ​ b = d ⟺ a d = b ( l o g a ​ b = l o g a ​ a d )

Answered by Anonymous | 2024-06-10

Log(6x/2)=1 Log(3x)=1 This is the same as saying 10^1=3x 10=3x x=10/3 or 3 and 1/3

Answered by HannahN | 2024-06-10

To solve lo g 6 x − lo g 2 = 1 , we simplify to lo g ( 3 x ) = 1 . From this we find 3 x = 10 , leading to the solution x = 3 10 ​ or approximately 3.33.
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Answered by Anonymous | 2024-12-24