Q 1 :

The number of solutions of  log4(x-1)=log2(x-3) is

  • 3

     

  • 1

     

  • 2

     

  • 0

     

(2)

For the given equation to be valid, we must have x-1>0 and x-3>0

 x>3

We can write the given equation as log(x-1)log4=log(x-3)log2

 log(x-1)=2log(x-3)       ( log4=2log2)

 (x-1)=(x-3)2x2-7x+10=0

 (x-5)(x-2)=0x=5 or 2

As x>3,  x=5



Q 2 :

The value of  log2log2log4256+2log22 is

  • 5

     

  • 2

     

  • 3

     

  • 4

     

(1)

log2log2log4256+2log22

=log2log2log444+2log2log2       [ logxx=1]

=log2log24+2log212log2

=log22+4=1+4=5



Q 3 :

Let n=2006! then 1log2n+1log3n++1log2006n=

  • 2006

     

  • 2005

     

  • 2005!

     

  • 1

     

(4)

1log2n+1log3n++1log2006n

=log2logn+log3logn++log(2006)logn

=1logn[log2+log3++log2006]

=1logn[log(1·2·32006)]

=1lognlog(2006!)=1logn×logn=1         ( n=2006!)



Q 4 :

If log35=a and log32=b, then log3300

  • 2(a+b)

     

  • 2(a+b+1)

     

  • 2(a+b+2)

     

  • a+b+4

     

(2)

log35=a,  log32=b

log35+log32=a+b,  log310=a+b

log3300=log3(3·100)=log33+log3(10)2

=log3(3)2+2log310=2+2(a+b)=2(a+b+1)



Q 5 :

The value of  log220log280-log25log2320 is equal to

  • 5

     

  • 6

     

  • 7

     

  • 8

     

(4)

log220log280-log25log2320

log220log280-log25log2(20×16)

=log220log2(16×5)-log25log2(20×16)

=log220(log216+log25)-log25(log220+log216)

=log220log216+log220log25-log25log220-log25log216

=log216(log220-log25)+0

=log216log2205=log216log24

=log224log222=4×2=8                ( log22=1)



Q 6 :

log4423(11024)=

  • - 3

     

  • 3

     

  • - 5

     

  • 5

     

(1)

We have,  log4423(11024)

=log(11024)log(4(42)1/3)=-log(2)10log45/3

=-log(2)10log[(2)2]5/3=-10log2103log2=-3



Q 7 :

The value of  6+log32(1324-1324-1324-132)

  • 4

     

  • 3

     

  • 2

     

  • 1

     

(1)

Let   t=4-1324-132

t2=4-132tt2+132t-4=0t=832

The required answer

=6+log32(132×832)=6+log3249=6-2=4



Q 8 :

The value of  ((log29)2)1log2(log29)×(7)1log47 is ______ .



(8)

((log29)2)1log2(log29)×(7)1log47

=(log29)2log2(log29)·(7)log74

=(log29)2loglog29(2)·712log74

=4·2=8



Q 9 :

The product of all positive real values of x satisfying the equation x(16(log5x)3-68log5x)=5-16 is _______ .



(1)

We have, x(16(log5x)3-68log5x)=5-16

Taking log both sides, we get

 t(16t3-68t)=-16                  [log5x=t]

 16t4-68t2+16=04t4-17t2+4=0

  t1+t2+t3+t4=0

 log5x1+log5x2+log5x3+log5x4=0x1x2x3x4=1