Problem

Problem

The system of equations log⁡10(2000xy)−log⁡10x⋅log⁡10y=4\log_{10}(2000xy) - \log_{10} x \cdot \log_{10} y = 4 log⁡10(2yz)−log⁡10y⋅log⁡10z=1\log_{10}(2yz) - \log_{10} y \cdot \log_{10} z = 1 and log⁡10(zx)−log⁡10z⋅log⁡10x=0\text{and } \log_{10}(zx) - \log_{10} z \cdot \log_{10} x = 0 has two solutions (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2) , then a. x1+x2=101x_1 + x_2 = 101 b. y1+y2=25y_1 + y_2 = 25 c. x1x2=100x_1 x_2 = 100 d. z1z2=100z_1 z_2 = 100
Intuitive explanation

This problem rewards substitution: let a=log⁡10xa = \log_{10} x, b=log⁡10yb = \log_{10} y, and c=log⁡10zc = \log_{10} z to convert the logarithmic system into polynomial equations in aa, bb, and cc. Once you expand each equation using log properties, you'll get three equations that are linear or quadratic in your new variables—much easier to handle than the original form. The key is recognizing that the system is actually hiding a simpler algebraic structure underneath the logarithms.

Intuitive explanation

This problem rewards substitution: let a=log⁡10xa = \log_{10} x, b=log⁡10yb = \log_{10} y, and c=log⁡10zc = \log_{10} z to convert the logarithmic system into polynomial equations in aa, bb, and cc. Once you expand each equation using log properties, you'll get three equations that are linear or quadratic in your new variables—much easier to handle than the original form. The key is recognizing that the system is actually hiding a simpler algebraic structure underneath the logarithms.

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