Question

Solve for x if a > 0 and 2 log_xa + log_axa + 3 log_a²x a = 0

(a) \(a^{\frac{3}{2}}\)

(b) \(a^{\frac{1}{2}}\)

(c) \(a^{\frac{3}{4}}\)

(d) \(a^{-\frac{4}{3}}\)

Answer 1

(d) \(a^{-\frac{4}{3}}\)

Since log_axa = \(\frac{1}{log_aax}\) = \(\frac{1}{log_aa+log_ax}\) = \(\frac{1}{1+log_ax}\) and

log_a²x a = \(\frac{1}{log_aa^2x}\) = \(\frac{1}{log_aa^2+log_{ax}x}\) = \(\frac{1}{2log_aa+log_{ax}x}\)

= \(\frac{1}{2+log_ax}\)

Given, 2 log_xa + log_axa + 3log_a²x a = 0

⇒ \(\frac{2}{log_ax}\) + \(\frac{1}{1+log_ax}\) + \(\frac{3}{2+log_ax}\) = 0

Now let log_ax = t, then \(\frac{2}{t}\) + \(\frac{1}{1+t}\) + \(\frac{3}{2+t}\) = 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 = \(-\frac{1}{2}\), \(-\frac{3}{4}\)

When t = \(-\frac{1}{2}\), log_ax = \(-\frac{1}{2}\) ⇒ \(x\) = \(a^{-\frac{1}{2}}\)

When t = \(-\frac{4}{3}\), log_ax = \(-\frac{4}{3}\) ⇒ \(x\) = \(a^{-\frac{4}{3}}\)