Page 32 - mathsvol1ch1to3ans

32

2
Comparing both sides, 7 = a + 10b 2
1
and 1 = ab ⇒ a =
b
1
Hence from first equation 7 = + 10b 2
b 2
2
7b = 1 + 10b 4
√
7 ± 49 − 40
Solving forb 2 =
20
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4 1
= or −
5 10
4
Since b is non negative b =
5
5
and a =
4
5 4 √
Hence the square root is = + 10
4 5

1 √
= (25 + 16 10)
20

Exercise - 2.17

x
1. Let b > 0 and b 6= 1. Express y = b in logarithmic form. Also state the domain and range of the
logarithmic function.
Solution: The logarithmic form is log y = x. The domain is (0, ∞) and the range is (−∞, ∞)
b
2. Compute log 27 − log 9 .
9
27
Solution:
3
log 27 − log 9 = log 2 3 − log 3 3 2
3
27
9
3
x
x
Using the expression log y a =
a
y
3 2
= −
2 3
5
=
6
3. Solve log x + log x + log x = 11.
4
2
8
1 1 1
Solution:Given + + = 11.
log 8 log 4 log 2
x
x
x
1 1 1
+ + = 11. Hence we have
3 log 2 2 log 2 log 2
x x x
1 1 1
+ + 1 = 11.
log 2 3 2
x
1 11
= 11.
log 2 6
x
1
6
= 6 log x = 6 ⇒ x = 2 = 64.
log 2 2
x