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Question

If sum of the solutions of the equation log2(32+x6x)=3+xlog2(32) is P, then

A
P is prime number
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B
log3P+logP3=2
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C
log3(P1)=0
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D
log3(P+1)=0
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Solution

The correct options are
B P is prime number
C log3P+logP3=2
log2(32+x6x)=3+xlog2(32)
log2⎜ ⎜32+x6x(32)x⎟ ⎟=3[mlogx=logxm&logalogb=logab]
23=32+x6x(32)x
8(32)x=9(3x)3x2x
(92x)2x=8
Put 2x=t
t29t+8=0
(t8)(t1)=0
t=8,1
2x=23,2x=1
x=3,x=0
Sum of solutions, P=3
Here, P is a prime number.
So, option A is correct.
Now,log3P+logp3=1+1=2
Option B is correct.
Also,log3(P1)=log320
log3(P+1)=log340
Option C and D are incorrect

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