Question

Solve for x if a > 0 and 2 logx a + logax a + 3 loga2x a = 0 -Maths 9th

Answer 1

(d) \(a^{-rac{4}{3}}\)Since logaxa = \(rac{1}{log_aax}\) = \(rac{1}{log_aa+log_ax}\) = \(rac{1}{1+log_ax}\) andloga2x a = \(rac{1}{log_aa^2x}\) = \(rac{1}{log_aa^2+log_{ax}x}\) = \(rac{1}{2log_aa+log_{ax}x}\)= \(rac{1}{2+log_ax}\)Given, 2 logxa + logaxa + 3loga2x a = 0⇒ \(rac{2}{log_ax}\) + \(rac{1}{1+log_ax}\) + \(rac{3}{2+log_ax}\) = 0Now let logax = t, then \(rac{2}{t}\) + \(rac{1}{1+t}\) + \(rac{3}{2+t}\) = 0⇒ 2(1 + t) (2 + t) + t(2 + t) + 3t (1 + t) = 0 ⇒ 2(2 + 2t + t + t2) + 2t + t2 + 3t + 3t2 = 0 ⇒ 4 + 6t + 2t2 + 2t + t2 + 3t + 3t2 = 0 ⇒ 6t2 + 11t + 4 = 0 ⇒ (3t + 4) (2t + 1) = 0 ⇒ t = \(-rac{1}{2}\), \(-rac{3}{4}\)When t = \(-rac{1}{2}\), logax = \(-rac{1}{2}\) ⇒ \(x\) = \(a^{-rac{1}{2}}\)When t = \(-rac{4}{3}\), logax = \(-rac{4}{3}\) ⇒ \(x\) = \(a^{-rac{4}{3}}\)