What is the answer to log_{10}(x)+log_{10}(x-1)=log_{10}(3x+12) ?

The answer to log_{10}(x)+log_{10}(x-1)=log_{10}(3x+12) is x=6

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log_{10}(x)+log_{10}(x-1)=log_{10}(3x+12)

lo g_{10} (x) + lo g_{10} (x - 1) = lo g_{10} (3 x + 12)

x = 6

Solution steps

lo g_{10} (x) + lo g_{10} (x - 1) = lo g_{10} (3 x + 12)

Apply log rules

x (x - 1) = 3 x + 12

Solve

x (x - 1) = 3 x + 12 : x = 6, x = - 2

x = 6, x = - 2

Verify Solutions:

x = 6

True

, x = - 2

False

The solution is

x = 6

lo g_{10} (x) + lo g_{10} (2) = lo g_{10} (5)

x, ln (y) = m ln (x) + ln (b)

x, y = ln (x)

lo g_{10} (2 x) = 3

x^{l o g_{10} (x)} = 100 x

What is the answer to log_{10}(x)+log_{10}(x-1)=log_{10}(3x+12) ?
The answer to log_{10}(x)+log_{10}(x-1)=log_{10}(3x+12) is x=6
Solve for x: log_{10}(x)+log_{10}(x-1)=log_{10}(3x+12)
The solution is x=6