1.Algebra Booster

Logarithm 3.7
fi
fi 10
log10x
log10x
- > 1
log103 log102
1 1
log x Ê
Á
- ˆ > 1
Ëlog 3 log 2˜
¯
fi log 10
x ¥ M > 1
fi log10
1
x >
M
10 10
fi x > 10 1/M
Hence, the solution set is x Π(10 1/M , )
1 1
EXAMPLE 6: Solve for x: - < 1.
log x log x - 1
Solution: Given in-equation is
2 2
È
Í
ÍÎ
2 2
1 1
- < 1
log x log x - 1
1 1
- < 1 where a = log x
a a - 1
fi 2
fi
fi
fi
fi
fi
fi
1 1
1 0
a
- - a - 1
<
1– a 1
0
a
- a - 1
<
2
–(1 – a)
- a < 0
aa ( –1)
2
(1 – a)
+ a > 0
aa ( –1)
2
a - a + 1 > 0
aa ( –1)
1
> 0
aa ( –1)
Ê 1 1 ˆ˘
M = Á - ˙
Ëlog103 log10
2˜
¯˙˚
fi a > 1 and a < 0
fi log 2
x > 1 and log 2
x < 0
fi x > 2 and x < 1
Also, log 2
x is defined only when x > 0.
Hence, the solution set is 0 < x < 1 and x > 2, i.e.
x Π(0, 1) (2, ).
EXAMPLE 7: Solve for x: log (2x+3)
x 2 < 1.
Solution: Given in-equation is
log (2x+3)
x 2 < 1
It is defined only when x π 0, 2x + 3 > 0, 2x + 3 π 1
fi x π 0, x > –3/2, x π –1
Case I: When 0 < 2x + 3 < 1
x 2 > 2x + 3
fi x 2 – 2x – 3 > 0
fi (x – 3)(x + 1) > 0
fi x < –1 and x > 3 …(i)
Also, 0 < 2x + 3 < 1
fi
3
- < x 1
x 2 < 2x + 3
fi x 2 – 2x – 3 < 0
fi (x – 3)(x + 1) < 0
fi –1 < x < 3
fi x Œ (–1, 3) …(iv)
Hence, the solution set [from Relations (iii) and (iv)] is
x
Ê 3 ˆ
Á 1
Ë
˜
2 ¯
( 1, 3) {0}
EXAMPLE 8: Solve for x:
Solution: Given in-equation is
fi
fi
fi
fi
2
log x - 3 log x + 3 < 1
log x - 1
2
2
log x- 3 log x+ 3 < 1.
log x - 1
a - 3a
+ 3 < 1 , where a = log x
a - 1
2
a - 3a
+ 3 - 1<
0
a - 1
2
a - 3a + 3– a + 1 < 0
a - 1
2
a - 4a
+ 4 < 0
a - 1
fi
1
< 0
a - 1
fi 0 < a < 1
fi 0 < log x < 1
fi 0 < log 10
x < 1
fi 1 < x < 10
Hence, the solution set is x Π(1, 10).
log 2( x + 1)
EXAMPLE 9: Solve for x: > 0.
( x - 1)
Solution: Given in-equation is
log 2( x + 1)
> 0
( x - 1)
It is possible only when x > 1, log 2
(x + 1) > 0 and x > –1.
Now, log 2
(x + 1) > 0
fi (x + 1) > 2 0 = 1
fi x > 0
Hence, the solution set is x Π(1, 10).