What is the answer to log_{4}(x^2+3x-1)=log_{4}(5x-1) ?

The answer to log_{4}(x^2+3x-1)=log_{4}(5x-1) is x=2

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log_{4}(x^2+3x-1)=log_{4}(5x-1)

lo g_{4} (x^{2} + 3 x - 1) = lo g_{4} (5 x - 1)

x = 2

Solution steps

lo g_{4} (x^{2} + 3 x - 1) = lo g_{4} (5 x - 1)

Apply log rules

x^{2} + 3 x - 1 = 5 x - 1

Solve

x^{2} + 3 x - 1 = 5 x - 1 : x = 2, x = 0

x = 2, x = 0

Verify Solutions:

x = 2

True

, x = 0

False

The solution is

x = 2

lo g_{x} (126) = 3

ln (2 x) = ln (x + 2)

lo g_{10} (10 x) = 3

lo g_{10} (10 x) = 2

x = \frac{lo g _{10} ( x ) + 7}{2}

What is the answer to log_{4}(x^2+3x-1)=log_{4}(5x-1) ?
The answer to log_{4}(x^2+3x-1)=log_{4}(5x-1) is x=2
Solve for x: log_{4}(x^2+3x-1)=log_{4}(5x-1)
The solution is x=2