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# Solve the following equation for x :2 logx a + logax a + 3 loga2x a = 0, a > 0

Navjyot Kalra
askIITians Faculty 654 Points
6 years ago
Sol. Given a > 0, so we have to consider two cases :
a ≠ 1 and a = 1. Also it is cleat that x > 0
And x ≠ 1, ax ≠ 1, a2x ≠ 1
Case I : If a > 0, ≠ 1
Then given equation can be simplified as
2/loga x + 1/loga x + 3/2 + loga x = 0
Putting loga x = y, we get
2(1 + y) (2 + y) + y(2 + y) + 3y(1 + y) = 0
⇒ 6y2 + 11y + 4 = 0 ⇒ y = -4/3 and -1/2
⇒ loga x = - 4/3 and loga x = -1/2
⇒ x = a-4/3 and x = a-1/2
Case II : If a = 1 then equation becomes
2 logx 1 + logx 1 + 3 logx­ 1 = 6 logx1 = 0
Which is true ∀ x > 0, ≠ 1
Hence solution is if a = 1, x > 0, ≠ 1
If a > 0, ≠ 1; x = a-1/2, a-4/3