The position of the Lagrange point L1
Let the distance from the Earth to the center of mass of the Earth-Moon system be r1, and the distance from the Earth to L1 be x. The distance between the Earth and the Moon is a. The mass of the Earth is M and of the Moon m. We assume that a body in L1 has a mass so small that the influence on the orbits of the Earth and the Moon is negligible.
The Earth, the Moon and a body in L1 are assumed to travel in circular orbits around the center of mass. The period of the moon is given by (derivation here):

In L1 the sum of the gravitational forces from the Earth and the Moon should be just sufficient to supply the required centripetal force to make a body in L1 go in a circular orbit with the same period as the Moon. The body should then continue to lie on a line connecting the Earth and the Moon. Newton's second law and gravitational law, and an expression for the centripetal acceleration, give
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We divide all terms by GM and multiply all terms by the common denominator. Putting k = m/M, we get:
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We know that (see the simulation "Orbit of the Moon")
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It is convenient to introduce the dimensionless quantity z = x/a. We divide the equation by a5 and use the expression for r1. We then get the following equation to determine the position of L1:
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By studying the graph of the lefthand side for various values of k, we discover that this fifth-order equation has one (and only one) solution in the interval 0 to 1.
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