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Aptitude Study Notes & Books-Free Download



Aptitude Round Question-Logarithms-Free Download


1.If x,a and m are any three numbers connected by the relation:

m=ax (a>0, a≠1), then,

"x" is defined as the logarithm of "m" to the base "a" and is written as:

x= loga m

2.Some important results:

(a) m =a log am

(b) x =log a (ax)

(c) log a 1 = 0

3.Some important theorems:

(a) log a (mn) = log a m + log a n

(b) log a (m/n) = log a m - log a n

(c) log a (mn) = n. log a m

(d) log a m = (log b m) / (log b a) ……. Change of base theorem

(e) log a a = 1

(f) log a b * log b a = 1

1.If ax = by, then

a.log a/b= x/y b.log a/ log b = x/y c.log a/ log b = y/x d.log b/a = x/y

2.2 log10 5 + log10 8 - ½ log10 4 = ?

a.2 b.4 c.2 + 2 log10 2 d.4 - 4 log10 2

3.loga (ab) = x, then log b (ab) is :

a.1/x b.x/ (x+1) c.x/(1-x) d.x/(x-1)

4.If log8 x + log 8 1/6 = 1/3, then the value of x is:

a.12 b.16 c.18 d.24

5.The value of (log9 27 + log8 32) is:

a.7/2 b.19/6 c.5/3 d.7

6.If log12 27 = a, then log6 16 is:

a.(3-a)/4(3+a) b.(3+a)/4(3-a) c.4(3+a)/(3-a) d.4(3-a)/(3+a)

7.The value of (1/log3 60 + 1/log4 60 + 1/log 5 60) is:

a.0 b.1 c.5 d.60

8.If log x + log y = log (x+y), then,

a.x=y b.xy=1 c.y= (x-1)/x d.y=x/(x-1)

9.If log 27= 1.431, then the value of log 9 is:

a.0.934 b.0.945 c.0.954 d.0.958

10.If log 2= 0.030103, the number of digits in 264 is :

a.18 b.19 c.20 d.21

Answer & Explanations

1.(c). ax = by => log ax = log b y => x log a = y log b

=> log a/ log b = y/x

2.(a). 2 log10 5 + log10 8 - ½ log10 4

= log10 (52) + log10 8 - log10 (41/2)

= log10 25 + log10 8 - log10 2 = log 10 (25*8)/2

= log10 100 = 2

3.(d). loga (ab) = x => log b/ log a = x => (log a + log b)/ log a = x

1+ (log b/ log a) = x => log b/ log a = x-1

log a/ log b = 1/ (x-1) => 1+ (log a/ log b) = 1 + 1/ (x-1)

(log b/ log b) + (log a/ log b) = x/ (x-1) => (log b + log a)/ log b = x/ (x-1)

=>log (ab)/ log b = x/(x-1) => logb (ab) = x/(x-1)

4.(a). log8 x + log8 (1/6) = 1/3

=> (log x/ log 8) + (log 1/6 / log 8) = log (8 1/3) = log 2

=> log x = log 2 - log 1/6 = log (2*6/1)= log 12

5.(c). Let log9 27 = x. Then, 9x = 27

=> (32)x = 33 => 2x = 3 => x= 3/2

Let log8 32 = y. Then

8y = 32 => (23)y = 25 => 3y = 5 => y=5/3

6.(d). log12 27 = a => log 27/ log 12 = a

=> log 33 / log (3 * 22) =a

=> 3 log 3 / log 3 + 2 log 2 = a => (log 3 + 2 log 2)/ 3 log 3 = 1/a

=> (log 3/ 3 log 3) + (2 log 2/ 3 log 3) = 1/3

=> (2 log 2)/ (3 log 3) = 1/a - 1/3 = (3-a)/ 3a

=> log 2/ log 3= (3-a)/3a => log 3 = (2a/3-a)log2

log16 16 = log 16/ log 6 = log 24/ log (2*3) = 4 log 2/ (log 2 + log 3)

= 4(3-a)/ (3+a)

7.(b). log60 3 + log60 4 + log60 5 + log 60 (3*4*5)

= log60 60 = 1

8.(d). log x + log y = log (x+y)

=> log (x+y) = log (xy)

=> x+y = xy => y(x-1) = x

=> y= x/(x-1)

9.(c). log 27 = 1.431 => log 33 = 1.431

=> 3 log 3= 1.431 => log 3 = 0.477

Therefore, log 9 = log 32 = 2 log 3 = (2*0.477) = 0.954

10.(c). log 264 = 64 log 2 = (64*0.30103) = 1926592

Its characteristics is 19.

Hence, the number of digits in 264 is 20



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