Solve each equation:

1) \(\log 5 - \log 2x = 1\)

2) \(2\log(x+1) = 5\)

0\\\\log(5:2x)=log10\\\\log\frac{5}{2x}=log10\iff\frac{5}{2x}=10\\\\20x=5\ \ \ /:20\\\\x=\frac{1}{4}\in D"> 1 ) l o g 5 − l o g 2 x = 1 ; D : x > 0 l o g ( 5 : 2 x ) = l o g 10 l o g 2 x 5 ​ = l o g 10 ⟺ 2 x 5 ​ = 10 20 x = 5 / : 20 x = 4 1 ​ ∈ D
0\to x > -1\\\\log(x+1)^2=log10^5\iff(x+1)^2=10^5\\\\x+1=-\sqrt{10^5}\ \vee\ x+1=\sqrt{10^5}\\\\x+1=-\sqrt{10^4\cdot10}\ \vee\ x+1=\sqrt{10^4\cdot10}\\\\x=-100\sqrt{10}-1\notin D\ \vee\ x=100\sqrt{10}-1\in D"> 2 ) 2 l o g ( x + 1 ) = 5 ; D : x + 1 > 0 → x > − 1 l o g ( x + 1 ) 2 = l o g 1 0 5 ⟺ ( x + 1 ) 2 = 1 0 5 x + 1 = − 1 0 5 ​ ∨ x + 1 = 1 0 5 ​ x + 1 = − 1 0 4 ⋅ 10 ​ ∨ x + 1 = 1 0 4 ⋅ 10 ​ x = − 100 10 ​ − 1 ∈ / D ∨ x = 100 10 ​ − 1 ∈ D

Answered by Anonymous | 2024-06-10

The first equation lo g 5 − lo g 2 x = 1 simplifies to x = 4 1 ​ after using logarithmic properties. The second equation 2 lo g ( x + 1 ) = 5 simplifies to x = 100 10 ​ − 1 . Both solutions are valid within their logarithmic constraints.
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Answered by Anonymous | 2024-12-24