Skip to content
MISCELLANEOUS EXERCISE 6 (I) · Q133

Q.The equation log⁡x216+log⁡2x64=3\log_{x^2}16+\log_{2x}64=3 has, (A) one irrational solution (B) no prime solution (C) two real solutions (D) one integral solution

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
63% · 133/210 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Let L=ln⁡2L=\ln2 and t=ln⁡xt=\ln x (so t≠0t\ne0, x≠1x\ne1; also need x>0, x≠1, 2x>0, 2x≠1x>0,\ x\ne1,\ 2x>0,\ 2x\ne1).

log⁡x216=ln⁡16ln⁡x2=4L2t=2Lt\log_{x^2}16=\dfrac{\ln16}{\ln x^2}=\dfrac{4L}{2t}=\dfrac{2L}{t}, and log⁡2x64=ln⁡64ln⁡(2x)=6LL+t\log_{2x}64=\dfrac{\ln64}{\ln(2x)}=\dfrac{6L}{L+t}.

The equation 2Lt+6LL+t=3\dfrac{2L}{t}+\dfrac{6L}{L+t}=3 clears to 2L2+5Lt−3t2=02L^2+5Lt-3t^2=0, a quadratic in LL: L=−5t±7t4L=\dfrac{-5t\pm7t}{4}, giving L=t2L=\dfrac{t}{2} or L=−3tL=-3t.

L=t2  ⟹  t=2L=2ln⁡2=ln⁡4  ⟹  x=4L=\dfrac{t}{2} \implies t=2L=2\ln2=\ln4 \implies x=4 (an integral solution). …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.