What is the answer to log_{10}(x)+log_{10}(2x-1)=log_{10}(6) ?

The answer to log_{10}(x)+log_{10}(2x-1)=log_{10}(6) is x=2

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log_{10}(x)+log_{10}(2x-1)=log_{10}(6)

lo g_{10} (x) + lo g_{10} (2 x - 1) = lo g_{10} (6)

x = 2

Solution steps

lo g_{10} (x) + lo g_{10} (2 x - 1) = lo g_{10} (6)

Apply log rules

x (2 x - 1) = 6

Solve

x (2 x - 1) = 6 : x = 2, x = - \frac{3}{2}

x = 2, x = - \frac{3}{2}

Verify Solutions:

x = 2

True

, x = - \frac{3}{2}

False

The solution is

x = 2

lo g_{10} (x) - lo g_{10} (2) = 1 + lo g_{10} (21 - x)

lo g_{3} (x) = - 1

lo g_{4} (x) + lo g_{10} (8 x) = 1

lo g_{10} (x + 1) = lo g_{10} (2 x + 1)

lo g_{10} (x + 2) - lo g_{10} (x - 1) = 4

What is the answer to log_{10}(x)+log_{10}(2x-1)=log_{10}(6) ?
The answer to log_{10}(x)+log_{10}(2x-1)=log_{10}(6) is x=2
Solve for x: log_{10}(x)+log_{10}(2x-1)=log_{10}(6)
The solution is x=2