Theory of Equations 1 Question 54

55. Solve 2logxa+logaxa+3logba=0, where a>0,b=a2x

(1978, 3M )

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Answer:

Correct Answer: 2. (x=a1/2 or x=a4/3)

Solution:

  1. The given equation can be rewritten as

2logax+1logaax+3logaa2x=0[b=a2x, given ]2logax+11+logax+32+logax=02t+11+t+32+t=0, where t=logax2(1+t)(2+t)+3t(1+t)+t(2+t)=06t2+11t+4=0(2t+1)(3t+4)=0t=12 or 43logax=12 or logax=43x=a1/2 or x=a4/3



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