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Identical expressions
- log(two)(x²- one)=log(two)(2x- one)
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Similar expressions
- log^(2)(2x-1)=3
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- log(2)(x²-1)=log(2)(2x+1)
log(2)(x²-1)=log(2)(2x-1) equation
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The solution
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/ 2 \ log(2)*\x - 1/ = log(2)*(2*x - 1)
$$\left(x^{2} - 1\right) \log{\left(2 \right)} = \left(2 x - 1\right) \log{\left(2 \right)}$$
Detail solution
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$\left(x^{2} - 1\right) \log{\left(2 \right)} = \left(2 x - 1\right) \log{\left(2 \right)}$$
to
$$- \left(2 x - 1\right) \log{\left(2 \right)} + \left(x^{2} - 1\right) \log{\left(2 \right)} = 0$$
Expand the expression in the equation
$$- \left(2 x - 1\right) \log{\left(2 \right)} + \left(x^{2} - 1\right) \log{\left(2 \right)} = 0$$
We get the quadratic equation
$$x^{2} \log{\left(2 \right)} - 2 x \log{\left(2 \right)} = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = \log{\left(2 \right)}$$
$$b = - 2 \log{\left(2 \right)}$$
$$c = 0$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 2$$
$$x_{2} = 0$$
left part with negative sign.
The equation is transformed from
$$\left(x^{2} - 1\right) \log{\left(2 \right)} = \left(2 x - 1\right) \log{\left(2 \right)}$$
to
$$- \left(2 x - 1\right) \log{\left(2 \right)} + \left(x^{2} - 1\right) \log{\left(2 \right)} = 0$$
Expand the expression in the equation
$$- \left(2 x - 1\right) \log{\left(2 \right)} + \left(x^{2} - 1\right) \log{\left(2 \right)} = 0$$
We get the quadratic equation
$$x^{2} \log{\left(2 \right)} - 2 x \log{\left(2 \right)} = 0$$
This equation is of the form
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = \log{\left(2 \right)}$$
$$b = - 2 \log{\left(2 \right)}$$
$$c = 0$$
, then
D = b^2 - 4 * a * c =
(-2*log(2))^2 - 4 * (log(2)) * (0) = 4*log(2)^2
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 2$$
$$x_{2} = 0$$