Question

Find x, if log_2x √x + log_2√x x = 0:

(a) 1, 2^–5/6 

(b) 1, 2^–6/5 

(c) 4, –2 

(d) None of these

Answer 1

(b) 1, 2^–6/5 

log_2x √\(x\) + log_2√\(x\) \(x\) = 0              ...(i)

Let log₂ \(x\) = t. Then,

 log_2x √\(x\) = \(\frac{log_2\,\sqrt{x}}{log_2\,2x}\) = \(\frac{\frac{1}{2}log_2\,x}{log_2\,2+log_2\,x}\) = \(\frac{\frac{t}{2}}{1+t}\)

 log_2x √\(x\) = \(\frac{log_2\,{x}}{log_2\,2\sqrt{x}}\) = \(\frac{log_2\,x}{log_2\,2+\frac{1}{2}log_2\,x}\) = \(\frac{t}{1+\frac{t}{2}}\)

∴ Substituting in (i), we get

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

⇒ t(2 + t) + 2t (2 + 2t) = 0 ⇒ 2t + t² + 4t + 4t² = 0 

⇒ 5t² + 6t = 0 ⇒ t(5t + 6) = 0 ⇒ t = 0 or – \(\frac{6}{5}\)

⇒ log₂ \(x\) = 0 ⇒ \(x\) = 20 = 1 and log₂\(x\) = – \(\frac{6}{5}\) ⇒ \(x\) = 2^-6/5

∴ \(x\) = 1 or 2^–6/5