1) lg (x-1)=3 (1 - lg5) 2)2Lg(x-2)-lg(3x-6)=2Lg2 3) lg(x+1)=1+lg12 4)2(lg5-1)=2Lg(x-3)-lg(3x+1)

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вопрос

1) To solve lg(x-1) = 3(1 - lg5), we can use the properties of logarithms to rewrite the equation as lg(x-1) = 3 - 3lg5. Now we can rewrite it in exponential form: 10^(3-3lg5) = x-1. Simplify the expression inside the parenthesis: 10^3 / 5^3 = x-1. Therefore, x = 15.

2) For 2lg(x-2) - lg(3x-6) = 2lg2, we can simplify the equation by using the properties of logarithms. Rewrite it as lg((x-2)^2) - lg(3x-6) = lg4. Combine the logarithms: lg((x-2)^2 / (3x-6)) = lg4. Convert it to exponential form: (x-2)^2 / (3x-6) = 4. Solve for x to get x = 6.

3) To solve lg(x+1) = 1 + lg12, rewrite it as lg(x+1) = lg12 + 1. This can be simplified to lg(12(x+1)) = 2. Change it back to exponential form: 12(x+1) = 10^2. Solve for x to get x = 81/12 = 6.75.

4) For 2(lg5-1) = 2lg(x-3) - lg(3x+1), simplify the equation using the properties of logarithms. Rewrite it as lg5^2 = lg((x-3)^2) - lg(3x+1). This simplifies to lg25 = lg((x-3)^2 / (3x+1)). Convert it to exponential form: 25 = (x-3)^2 / (3x+1). Solve for x to get x = 4.