1.Algebra Booster

3.20 Algebra Booster
fi
12¥
3
4
( xy) = 2 = 2 9
fi 2xy = 2 10 = 1024
14. We have,
log 10
(x 2 + x) = log 10
(x 3 – x)
fi x + 1 = x 2 – 1
fi x + 1 = (x + 1)(x – 1)
fi x – 1 = 1
fi x = 2
Hence, the solution is 2.
Thus, the product of all the solutions = 2.
15. We have log 10
(x – 2) + log 10
y = 0
fi log 10
[y(x – 2)] = log 10
1
fi y(x – 2) =1 …(i)
Also, x + y - 2 = x+
y
fi x+ y - 2+ 2 x( y - 2) = x+
y
fi - 2+ 2 xy ( - 2) = 0
fi - 2= -2 xy ( -2)
fi x(y – 2) = 1 …(ii)
From Eqs (i) and (ii), we get
x = y
Put x = y in Eq. (ii), we get
x(x – 2) = 1
fi x 2 – 2x – 1 = 0
fi x 2 – 2x + 1 = 2
fi (x – 1) 2 = 2
fi x –1=±
2
fi x = 1±
2
fi x= 1+ 2 = y
( x = y)
Thus, the value of
x+ y -2 2
= 1+ 2 + 1+ 2 -2 2
= 2
16. We have,
7
log27a
+ log9b=
2
fi
7
log 3a
+ log 2b=
3 3 2
fi 1 log
1 7
3a
+ log3b=
3 2 2
fi 2 log 3
a + 3 log 3
b = 21 …(i)
2
Also, log27b+ log9a=
3
fi log
1 2
3b+ log3a
=
3 2 3
fi 2 log 3
b + 3 log 3
a = 4 …(ii)
Solving Eqs (i) and (ii), we get
4 log 3
a – 9 log 3
a = 42 – 12
fi –5 log 3
a = 30
fi log 3
a = –6
fi a = 3 –6
From Eq. (ii), we get
2 log 3
b – 18 = 5
fi 2 log 3
b = 2
fi log 3
b = 1
fi b = 3 11
Hence, the value of ab = 3 –6 ¥ 3 11 = 243.
17. Given equation is
log tan x
(2 + 4 cos 2 x) = 2
fi
fi
2 + 4 cos 2 x = tan 2 x
2 sin x
2+ 4cos x =
2
cos x
2
fi 4 cos 4 x + 2 cos 2 x – sin 2 x = 0
fi 4 cos 4 x + 3 cos 2 x – 1 = 0
fi 4 cos 4 x + 2 cos 2 x – cos 2 x – 1 = 0
fi 4 cos 2 x (cos 2 x + 1) – (cos 2 x + 1) = 0
fi (4 cos 2 x – 1)(cos 2 x + 1) = 0
fi (4 cos 2 x Р1) = 0 ( x Π[0, 2p)
2
2 1 2
Ê ˆ Êp
ˆ
fi cos x = Á = cos
Ë
˜ Á ˜
2¯ Ë 3¯
p
fi x = np
± , n = 0,1,2
3
p 2p 4p 5p
fi x = , , ,
3 3 3 3
Hence, the number of values of x is 4.
18. Given,
cos (ln x) = 0
fi
p
ln x =
2
fi x=
e p
2
Thus, the value of
Ê 2
ˆ
Á ¥ log ( x) + 10
Ë
˜
p
¯
2
p
Ê
log(
2
ˆ
= Á ¥ e ) + 10
Ë
˜
p
¯
Ê 2 p ˆ
= Á ¥ log ( e) + 10
Ë
˜
p 2 ¯
= 1 + 10 = 11
19. Given, c(a – b) = a(b – c)
fi ac – bc = ab – ac
fi 2ac = ab + bc = b(a + c)
fi
2ac
b =
( a + c)