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Question

The equation (1+logx27)log3x+1=0 has

A
no integral solution
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B
one irrational solution
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C
two real solutions
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D
no prime solution
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Solution

The correct option is D no prime solution
Clearly, x>0,x1
(1+logx27)log3x+1=0
(1+32logx3)log3x+1=01+32log3x=1log3x

Let log3x=t
Then 1+32t=1t, t0
1+32t=1t22t2+3t2=0t=2,12log3x=2,12x=19,3

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