What is the answer to ln(2x+1)=ln(x^2+x-2) ?

The answer to ln(2x+1)=ln(x^2+x-2) is x=(1+sqrt(13))/2

Solve for x: ln(2x+1)=ln(x^2+x-2)

The solution is x=(1+sqrt(13))/2

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ln (2 x + 1) = ln (x^{2} + x - 2)

x = \frac{1 + 13}{2}

Decimal Notation

x = 2.30277 \dots

Solution steps

ln (2 x + 1) = ln (x^{2} + x - 2)

Apply log rules

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

Solve

2 x + 1 = x^{2} + x - 2 : x = \frac{1 + 13}{2}, x = \frac{1 - 13}{2}

x = \frac{1 + 13}{2}, x = \frac{1 - 13}{2}

Verify Solutions:

x = \frac{1 + 13}{2}

True

, x = \frac{1 - 13}{2}

False

The solution is

x = \frac{1 + 13}{2}

lo g_{2} (5 r - 2) = 1

lo g_{5} (3 x) + 3 lo g_{5} (2) = 3

- lo g_{10} (x) = 13.5

x, y^{3} = ln (x y)

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

What is the answer to ln(2x+1)=ln(x^2+x-2) ?
The answer to ln(2x+1)=ln(x^2+x-2) is x=(1+sqrt(13))/2
Solve for x: ln(2x+1)=ln(x^2+x-2)
The solution is x=(1+sqrt(13))/2