1.Algebra Booster

Logarithm 3.19
Again, log a
b = 2
fi log 3
b = 2
fi b = 3 2 = 9
Further, log b
c = 2
fi log 9
c = 2
fi c = 9 2 = 81
Thus, (a + b + c) + 7 = 3 + 9 + 81 + 7 = 100
8. We have,
7
log9x+ log4y
=
2
fi 1 log
1 7
3x+ log2y
=
2 2 2
fi log 3
x + log 2
y = 7 …(i)
3
Also, log9x
- log8y
= -
2
fi log 3
x – log 2
y = –3 …(ii)
Adding Eqs (i) and (ii), we get
2 log 3
x = 4
fi log 3
x = 2
fi x = 3 2 = 9
Subtracting Eqs (i) and (ii), we get
2 log 2
y = 10
fi log 2
y = 5
fi y = 2 5 = 32
Hence, the solutions are x = 9 and y = 32.
9. Given 3 x+2 = 45
fi x + 2 = log 3
(45)
= log 3
(5 ¥ 9)
= log 3
5 + log 3
9
= log 3
5 + 2
fi x = log 3
5
fi
log105
x =
log10
3
fi
Ê10ˆ
log10
Á
Ë
˜
2 ¯ log1010 - log10
2
x = =
log10 3 log10
3
fi
1 - a
x =
b
10. We have,
N = 6 log 10
2 + log 10
31
= log 10
2 6 + log 10
31
= 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.
11. We have
2 Ê1ˆ
-2 2
M = log (log (2 ))
2 Á =
Ë
˜ 2
4¯
2
Ê 2 ˆ
= Á-
log 2 2
Ë
˜
1/2 ¯ = (–4)2 = 16
3 3
and N = log (8) = [log (8)]
2 2 2 2
3 3 3
= [log (2 2) ] = [3 log (2 2)]
2 2 2 2
3
= (3) = 27
Also,
5
P = log 5(log 3( 9 ))
= log 5
[log 3
(9 1/10 )]
= log 5
[log 3
(3 2/10 )]
Ê1ˆ
= log 5Á [log 3(3)]
Ë
˜
5¯
= log 5
(5 –1 ) = –1
ÊM
ˆ 16
Thus, Á + P + 3
Ë
˜ = - 1+
3
N ¯ 27
16
= + 2
27
70
=
27
12. We have,
1 1
+ = log
a b x
3 + log x
7
= log x
(21)
1
Thus, log 21( x)
=
log x(21)
1 ab
= =
1 1
+
a + b
a b
13. We have,
log 8
x + log 4
y 2 = 5
fi 1 log 2
2x+ log2y
= 5
3 2
fi log 2
x 1/3 + log 2
y = 5
fi log 2
(x 1/3 y) = 5
fi (x 1/3 y) = 2 5 = 32 …(i)
Also, log 8
y + log 4
x 2 = 7
1 log + 2 log x=
7
3 2
fi 2y
2
fi log 2
y 1/3 + log 2
x = 7
fi log 2
(y 1/3 x) = 7
fi (y 1/3 x) = 2 7 = 128 …(ii)
Multiplying Eqs (i) and (ii), we get
(x 4/3 y 4/3 ) = 27 =128
fi (xy) 4/3 = 2 12