solve the logarithmic equation ....round to 3 decimal places?

Question

Log√6-x - Log√x-4 = Log x

Answer 1

sqrt[(6-x)/(x-4)] = x

(6-x)/(x-4) = x^2
6-x = x^3 - 4x^2
x^3 - 4x^2 +x - 6 =0

Answer 2

I am interpreting this as

Log[√(6-x)] - Log[√(x-4)] = Log x

Note that, in order for these to be real-valued, we must have 6 - x ≥ 0 and x - 4 ≥ 0, and furthermore we cannot have x = 4. So any real-valued solution must satisfy 4 < x ≤ 6.

Use the fact that Log A - Log B = Log (A/B) to combine the terms on the left side:

Log[√(6-x)/√(x-4)] = Log x

Exponentiate both sides, using as base whatever value the base of the logarithms is (10, e, etc.):

√(6-x)/√(x-4) = x

Square both sides, realizing that this can introduce spurious solutions:

(6 - x)/(x - 4) = x²

Multiply both sides by (x - 4) and multiply out the right side:

6 - x = x³ - 4x²

0 = x³ - 4x² + x - 6

This cubic has a single real root at x ≈ 4.11169; note that this value does lie in the required interval 4 < x ≤ 6. More importantly, it satisfies (to within roundoff error) the original equation, so the squaring done during the solution process did not introduce a spurious result.