1.Algebra Booster

Logarithm 3.29
= log 10
(64 ¥ 31)
= log 10
(1984)
< log 10
(1000) = 3
Also, N = log 10
(1984) > log 10
(10000) = 4
Thus, the sum of successive integers = 3 + 4 = 7.
10. Let S = 1 + 1 + 1 + … = 1/3 =
1
2 3
3 3 3 1 - (1/3) 2
We have
(0.16)
1
1 log 5
( ) Ê 4 ˆ
( 2)
log2.5 2
2
= Á
Ë ˜
25¯
Ê2ˆ
= Á
Ë ˜
5¯
Ê2ˆ
= Á
Ë ˜
5¯
1
( )
2log5 2
2
1
( )
1
( )
–2log 5 2
2
-2
log 5 -2
2
2
Ê2ˆ Ê1ˆ
= Á = = 4
Ë
˜ Á ˜
5¯ Ë2¯
11. We have,
ab = log 12
18 ¥ log 24
54
log 18 log 54
= ¥
log 12 log 24
2 3
log(2.3 ) log(2.3 )
= ¥
2 3
log(3.2 ) log(3.2 )
log 2 + 2log 3 log 2 + 3 log 3
= ¥
log(3) + 2log 2 log 3 + 3 log 2
1 + 2 log23 1 + 3 log23
= ¥
log 2(3) + 2 log23 + 3
1+ 2x
1+
3 x
= ¥ , x = log 2 3
x+ 2 x+
3
2
1+ 5x+
6x
=
2
x + 5x+
6
Also,
(a – b) = log 12
18 – log 24
54
2 3
log (2 ¥ 3 ) log (2 ¥ 3 )
= -
2 3
log (3 ¥ 2 ) log (3 ¥ 2 )
log 2 + 2 log 3 log 2 + 3 log 3
= -
log (3) + 2 log 2 log 3 + 3 log 2
1 + 2 log23 1 + 3 log23
= -
log 2(3) + 2 log23 + 3
1+ 2x
1+
3x
= - where x = log 2 3
x + 2 x + 3
2
(1 - x )
= (2 + x )(3 + x )
Hence, the value of
2 2
5(1 - x ) (1 + 5x+
6 x )
5(a – b) + ab = +
(2 + x)(3 + x) (2 + x)(3 + x)
2 2
5(1 - x ) + (1 + 5x+
6 x )
=
(2 + x)(3 + x)
2
x + 5x+
6
=
(2 + x)(3 + x)
( x+ 2)( x+
3)
=
(2 + x)(3 + x)
= 1
Previous Years’ JEE-Advanced Examinations
1. The given equation is
fi
fi
2log xa + logaxa + 3 log 2 a = 0
a x
2log a log a 3 log a
+ + = 0
2
log x log ( ax) log ( ax)
2 1 3
+ + = 0
log x log a + log x 2log a + log x
fi 2 + 1 + 3 = 0
y b+ y 2b+
y
where log a = b, log x = y
fi 6y 2 + 11by + 4b 2 = 0
fi 6y 2 + 3by + 8by + 4b 2 = 0
fi (2y + b)(3y + 4b) = 0
fi
b 4b
y =- , -
2 3
b
When y =- ,
2
log a
log x = – 2
2 1
fi
Ê ˆ
log x =- log a= log Á
Ë
˜
a¯
fi
x
2 1
=
a
fi x = a –1/2
4b
When y =- ,
3
4loga
log x =-
3
fi
fi
fi
Ê1ˆ
3 log x = 4 logÁ Ë
˜
a¯
log x
x
3 Ê1
3 Ê1
ˆ
= Á
Ë
˜
a¯
fi x = a –4/3
ˆ
= log Á
Ë
˜
a¯
4
4