If (5.55)x = (0.555)y = 1000, then the value of 1/x − 1/y is:

Concept: For an exponential equation of the form am = N, taking log10 of both sides gives m·log10(a) = log10(N), so 1/m = log10(a) / log10(N). When two different bases raised to exponents both equal the same N, subtracting the reciprocals of the exponents converts the ratio of the bases into a single logarithm.

Application:

1. From (5.55)x = 1000, take log10 of both sides: x·log10(5.55) = log10(1000) = 3, so 1/x = log10(5.55) / 3.

2. From (0.555)y = 1000, similarly: y·log10(0.555) = 3, so 1/y = log10(0.555) / 3.

3. Subtract: 1/x − 1/y = [log10(5.55) − log10(0.555)] / 3 = log10(5.55 / 0.555) / 3.

4. Compute the ratio of the two bases: 5.55 / 0.555 = 10.

5. So 1/x − 1/y = log10(10) / 3 = 1/3, since log10(10) = 1.

Cross-check: Write 5.55 = 1000(1/x) and 0.555 = 1000(1/y). Dividing gives 5.55 / 0.555 = 1000(1/x − 1/y), i.e. 10 = 1000(1/x − 1/y). Since 1000(1/3) = 10 (the cube root of 1000 is 10), matching exponents confirms 1/x − 1/y = 1/3 — the same result as above.

Result: 1/x − 1/y = 1/3.