# Logarithm MCQ

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Logarithm MCQ

1. Which one is correct?

(a)    log (m + n) = log m + log n

(b)    log (m – n) = log m – log n

(c)    logb a x loga b =1

(d)    None of these.

logb a x loga b = loga a x logb b = 1 x 1 = 1. So, option (c) is correct

2. Find the value of log b a . logc b . loga c

(a)    abc

(b)    0

(c)    1

(d)    4.

logb a. logc b. loga c = (logc a. logb b). loga c = (logc a x 1) loga c = logc a  x  loga c = loga a. logc c = 1 x 1 = 1. So, option (c) is correct

3. If log 2 = 0.30103, then the value of log 200 is

(a)    18.0103

(b)    6.30103

(c)    4.30103

(d)    2.30103

log 200 = log (2 x 100) = log 2  +  log 100 = log 2  +  2 log 10 = 0.30103 + 2 X 1 = 2.30103

(we know log 2 = 0.30103, log 10 = 1). So, option (d) is correct

4. log 144 is equal to

(a)    2 log 4 + 2 log 2

(b)    4 log 2 + 2 log 3

(c)    3 log 2 + 4 log 3

(d)    None of these.

• 2 log 4 + 2 log 2 = log 42 + log 22 = log 16 + log 4 = log (16 x 4) = log 64
• 4 log 2 + 2 log 3 = log 24 + log 32 = log 16 + log 9 = log (16 x 9) = log 144
• 3 log 2 + 4 log 3 = log 23 + log 34 = log 8 + log 81 = log (8 x 81) = log 648

So, option (b) is correct

5. The relation loge x + loge (1 + x) = 0 is equal to

(a)    x2 + x + e = 2

(b)    x2 + x – e = 3

(c)    x2 + x – 1 = 0

(d)    x2 + x + 4 = 0.

loge x + loge (1 + x) = 0, Or,  loge x(1 + x) = 0, Or, x(1 + x) = e0, Or, x + x2  = 1, Or, x2 + x – 1 = 0. So, option (c) is correct

6. Find value of  (½)log10 (25) – 2 log10 (3) +log10 (18)

(a)      0

(b)     1

(c)      3

(d)     2. log10[an1]  25 – 2 log10  3 + log10 18= log 10 25 ½ –  log 10 32 +log 10 18 = log 10 5 –  log 10 9 +log 10 18

(log 10 5 –  log 10 9) + log 10 18 = log 10 (5/9) + log 10 18 = log 10 [5 X 18)/9] = log 10 10 = 1.

So, option (b) is correct

7. If log10 2 = 0.3010, the value of log10 80 is

(a)    1.9030

(b)    1.6020

(c)    2.9030

(d)    4.9030.

Log10 80=log10 (8 X 10) = log10  8  + log10 10 = log10  23 + 1 = 310 log  2 + 1

= 3 x 0.3010 + 1 = 0.9030 + 1 = 1.9030. So, option (a) is correct

8. Find the value of log3 (27)3

(a)    27

(b)    9

(c)    729

(d)    81.

log3 (27)3 = 3 log3 (27)= 3 log3 33    = (3 X 3) log3 3  = 9 log33 = 9  X1 = 9 (we know log3 3 = 1).

So, option (b) is correct

9. If log3 (log2 x) = 1, Find the value of x

(a)    10

(b)    9

(c)    3

(d)    8

log3 (log2 x) = 1, or, log3 (log2 x) = 1, Now, log2 x = 31 = 3, Or, x = 23 = 8, So, option (d) is correct

10. If log4 (x2 + x) log4 (x + 1) = 2, then:

(a)    x = 8

(b)    x = 4

(c)    x = 6

(d)    x = 16.

log4 (x2 + x) – log4 (x + 1) = 2,  or, log4 [(x2 + x) / (x + 1)] =2. Or, [(x2 + x) / (x + 1)] = 42 = 16

Or, x2+ x = 16(x + 1), Or, x2+ x = 16x + 16, Or, x2+ x – 16x – 16 = 0 , Or, x(x +1) – 16(x +1) = 0

Or, (x +1) (x -16), Either   x = -1  or,  x = 16, So, option (d) is correct

11. Value of  (log10 125 / log10 25) = ?

a. 5

b. $\displaystyle \frac{3}{2}.$

c. $\displaystyle \frac{1}{2}$

d. $\displaystyle \frac{1}{6}$

(log10 125 / log10 25) = (log10 53 / log10 52) = 3 log10 5 / 2 log10 5 = 3/2. So, option (b) is correct

12. If log8 m + log8 2=2/3, then find the value of m

(a)    1

(b)    3

(c)    2

(d)    4

log8 m + log8 2=2/3, or, log8 (m X 2) + =2/3, or, log8 2m =2/3,

So, 8 (2/3) = 2m. or, {23}(2/3) = 2m. Or, 22 = 2m, Or, m = 2. So, option (c) is correct

13. Value of 7 log10 (16/15)+ 5 log10 (25/24)+ 3 log10 (81/80)

(a)    2

(b)    log10 2

(c)    log10 3

(d)    log10 5

7 log10 (16/15)+ 5 log10 (25/24)+ 3 log10 (81/80)

= 7 {log10 16 –  log1015)} + 5 {log10 25 – log10  24} + 3 [{ (log10  34) – log10 (80}]

= 7 {log10 24 –  log10 (5 X 3)} + 5 {log10 52 – log10 (8X3)} } + 3 { log10  (34)- log10 (24 X 5)}

= 7 { 4log10 2 – ( log10 5 + log10 3)} + 5 {2 log10 5- (log10 23 + log103) } + 3 { 4log10  3  – (4log10 2 + log10 5)}

= 28 log10  2 – 7 log10  5- 7 log10 3 +10 log10  5 – 15 log10 2 – 5 log10 3 + 12 log10  3 – 12 log 10 2- 3 log 10 5

= log 10 2. So, option (b) is correct

14. If log10 p + log10q = log10 (p + q), which one of the following relations is true?

(a)    p=(q/q-1)

(b)    3p=q=0

(c)    3p=q2/(1-q)

(d)    p=q=2

log10   p + log10  q = log10  (p + q), Or,  log10   (pq) = log10  (p + q), Or,  pq = p + q, Or,  pq – p = q,

Or,  p(q – 1) = q, or, p=(q/q-1). So, option (a) is correct

15. If log [(x+y)/5] = ½ (logx + log y), find value of  (x/y) + y/x)

(a)       20

(b)       23

(c)       25

(d)       18.

log [(x+y)/5] = ½ (logx + log y), or, log [(x+y)/5] = ½ log (xy), or,  log [(x+y)/5] = ½ log (xy),

log [(x+y)/5] = log [(xy) ½ ].  Or, (x+y)/5 = (xy) ½, or {(x+y)/5}2 = (xy), or, (x2 + y2 + 2xy) / 25=xy

or, (x2 + y2 + 2xy) =25xy,  x2 + y2=23xy, or, (x2 + y2) / xy =23,

Now, (x/y) + y/x) = (x2 + y2) / xy = 23. So, option (b) is correct

16. If log (2a 3b) = log a log b, then a =

(a)       3b² / (2b – 1)

(b)       3b/  (2b – 1)

(c)       b²/[4 (2b + 1)]

(d)       None of the above

log (2a – 3b) = log a – log b, or, log (2a – 3b) = log (a/b). or, 2a-3b=a/b, Or, 2ab – 3b2 = a

Or, 2ab – a = 3b2, Or, a(2b – 1) = 3b2, or a= 3b2/ (2b – 1). So, option (a) is correct

17. Find value of log (a2/bc) + log (b2/ac) + log (c2/ab)

(a)    1

(b)    0

(c)    3

(d)    log a

log (a2/bc) + log (b2/ac) + log (c2/ab)

= (log a2 – log bc) + (log b2 – log ac) + (log c2 – log ab)

= log a2 + log b2 + log c2 – (log bc + log ac + log ab)

= log a2 b2 c2 – log (bc. .ab)

= log a2 b2 c– log a2 b2 c2  = 0. So, option (b) is correct

18. If log (m/n) + log (n/m) = log (m+n) then find true statement

(a)      m+n=1

(b)     m/n=2

(c)      m – n = 2

(d)     m2 – n2 = 4

log (m/n) + log (n/m) = log (m+n), or log {(m/n) X (n/m)} = log (m+n), or log1=log (m+n). or m+n=1. So, option (a) is correct

19. If loga3= 1/3, then find value of a

(a)    27

(b)    81

(c)    9

(d)    24.

loga3= 1/3, So, a1/3=3, or a=(3)3 = 27. So, option (a) is correct

20. Find the value of log2 (1/64)

(a)    6

(b)    -6

(c)    64

(d)    4.

log2 (1/64) = log2 (1/26) =  log2 1 – 6log2 2 = 0 – (6X 1) = -6 (as  log2 2= 1). So, option (b) is correct

21. logx(16/25) = -1/2, Find value of x

(a)    625/256

(b)    256/625

(c)    526/265

(d)    32/50

logx(16/25)=-1/2. Or x(-1/2) =16/25, or  1/{x(1/2)} =16/25, or x(1/2)} = 25/16, or x=252/162= 625/256. So, option (a) is correct

22. If logx10000 = -1/4, find x

logx10000 = -1/4, so, (10000)-1/4= x, or 1/ (10000)1/4 = x, or x= 1/{(104)}1/4 = 1/10.

So, option (a) is correct

23. Find value of log10 (0.00001)

(a)      – 5

(b)     1/5

(c)      – 4

(d)     5

log10 (0.00001) = log10 (1/100000) = log10 (1/105) = log10 10-5) = 5 log10 = -5 X1= – 5 .

So, option (a) is correct

24. If log10 x + log10y = z, then find value of x

(a)    z/y

(b)    10/3z

(c)    10z/y

(d)    yz

log10   x + log10  y = z, or Or, log10   (xy) = z,  10z = xy, x=10z/y. So, option (c) is correct

25. Find value of log10 5 assuming log10 2 = 0.3010

(a)    .4010

(b)    0.6911

(c)    0.6990

(d)    .6021.

log10  5 = log10 (10/2) = log10 10 – log10 2 = 1- 0.3010 =0.6990. So, option (c) is correct

26. Find value of (log5 3) x (log3 625)

(a)    1

(b)    2

(c)    3

(d)    4.

log3 x log3  625 = log 3 3 x log5  625 (by change of base) = 1 x log5  54 = log5  54 = 4 log5 5 = 4X1=4. So, option (d) is correct

27. If log(x2 6x + 10) = 0, then the value of x is

(a)    4

(b)    7

(c)    3

(d)    5.

log  (x2 – 6x + 10) = 0, Or, log( x2 – 6x + 10) = log 1, x2 – 6x + 10=1, or  x2 – 6x + 10 – 1 = 0

Or, x2 – 6x + 9 = 0, Or, x2 – 3x – 3x + 9 = 0, Or, x (x – 3) – 3(x – 3) = 0, Or, (x – 3) (x – 3) = 0, or x = 3. So, option (c) is correct

28. Find x, if log43 + log4(x + 2) = 2

(a)    10/3

(b)    4/3

(c)    2/3

(d)    12/5

log4   3 + log4 (x + 2) = 2,  log4   {3 X (x+2)} =2, or, log4   {3x+6)} =2,

Or, 42 = 3x + 6  Or, 3x+ 6 = 16 Or, 3x = 16- 6 = 10 Or, x = 10/3. So, option (a) is correct

29.  If a2 + b2 = 23ab, evaluate log {(a+b)/5}

(a)    5

(b)    log(ab) /2

(c)    log(ab) /5

(d)    log(a+b) /4

a2 + b2 = 23ab, Or, a2 + b2 + 2ab = 23ab + 2ab, Or, (a + b)2 = 25ab, {(a+b)/5)}2=ab, or (a+b/5)=(ab) ½

or log {(a+b/5)}=log {(ab) ½ } = ½ log ab = {log(ab)} /2. So, option (b) is correct

30.Evaluate {1/logab(abc)} + {1/logbc(abc)} + {1/logca(abc)}

{1/logab(abc)} + {1/logbc(abc)} + {1/logca(abc)}

= 2  log abc abc = 2.1 = 2. So, option (c) is correct

31. If log (2a – 3b) = log a – log b, find value of a

(a)    3b2/(2b – 1)

(b)    3b/2b – 1

(c)    2b2/(4b + 2)

(d)    2b2/2b – 1

log (2a – 3b) = log a – log b, Or, log (2a – 36) = log (a/b), Or, 2a – 3b = a/b

Or, 2ab – 3b2 = a, Or, 2ab – a = 3b2, Or, a(2b – 1) = 3b2, or a=3b2 / (2b – 1). So, option (a) is correct

32. If log (a+b/4) = 1/2 (log a + log b), then find value of (a/b + b/a)

log (a+b/4) = 1/2 (log a + log b), or log (a+b/4) = log (ab)1/2, or, (a+b/4) =(ab)1/2,