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Transcript
17
ACC-MT- LOGARITHM
Solved Examples
Q.1
Sol.
Find the value of x satisfying log10 (2x + x – 41) = x (1 – log105).
We have,
log10 (2x + x – 41) = x (1 – log105)

log10(2x + x – 41) = x log102 = log10 (2x)

2x + x – 41 = 2x  x = 41. Ans.
Q.2
If the product of the roots of the equation,
Sol.
(where a, b  N) then the value of (a + b).
Take log on both the sides with base 2
3
5
2
  (log2 x )  log 2 x   
4

4
x
 2 is
1
a
b
5
3
1
2
 log 2 x   log 2 x   log 2 x =
4
4
2


log2x = y
3y3 + 4y2 – 5y – 2 = 0 
3y2(y – 1) + 7y(y – 1) + 2(y – 1) = 0

(y – 1)(3y2 + 7y + 2) = 0

(y – 1)(3y + 1)(y + 2) = 0
1

y = 1 or y = – 2 or y =
3
1
1
1

x = 2; ; 1 3  x1x2x3 = 3
 a + b = 19
4
2
16
Q.3
For 0 < a  1, find the number of ordered pair (x, y) satisfying the equation loga 2 x  y =
loga y  loga x  log 2 4 .
a
Sol.
We have loga 2 x  y =
1
2

 y 
Also, log a 
  log 2 4 
a
| x |
a
2a
If x > 0, then x = , y =
3
3
If x < 0, then y = 2a, x = – a

|x+y|=a

y=2|x|
.....(2)
x+y=±a
 a 2a 
 and (– a, 2a)
possible ordered pairs =  ,
3 3 
Q.4
The system of equations
log10(2000xy) – log10x · log10y = 4
log10(2yz) – log10y · log10z = 1
and
log10(zx) – log10z · log10x = 0
has two solutions (x1, y1, z1) and (x2, y2, z2). Find (y1 + y2).
Sol.
From (1),
3 + log10(2xy) – log10x · log10y = 4
or
log10(xy) – log10x · log10y = 1 – log10(2)
....(i)
BANSAL CLASSES Private Ltd. ‘Gaurav Tower’, A-10, Road No.-1, I.P.I.A., Kota-05
.....(1)
1
and
2