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Indices MCQ

1. If 3^x = 4^y = 12^z, then z is equal to:

a. (x+y)/x

b. xy/(x+y)

c. y/(x+y)

d. (x+y)/y

Let 3^x = 4^y = 12^z = k. So, 3=k^1/x, 4=k^1/y, 12=k^1/z.

So, k^1/x X k^1/y = 3X 4=12. Or k^{1/x + 1/y} = 12. k^1/z =12 = k ^{(1/x + 1/y)}. So 1/z= 1/x + 1/y, or 1/z = (x+y)/xy.

Or z=xy/(x+y). So, option (b) is correct.

2. (x^b+c)^b–c (x^c+a)^c–a (x^a+b)^a–b is equal to:

(a)      0

(b)      1

(c)      ab

(d)      ac

(x^b+c)^b–c  (x^c+a)^c–a  (x^a+b)^a–b = x ^{b²– c²}  .  x ^{c²– a²}  .  x ^{a²– b²} =  x⁰ =1. So, option (b) is correct.

3. The value of (x^a/x^b)^a+b X (x^b/x^c)^b+c X (x^c/x^a)^c+a is

(a)    1

(b)    0

(c)    x

(d)    x ^{a + b}

(x^a/x^b)^a+b X (x^b/x^c)^b+c X (x^c/x^a)^c+a =   (x^{a– b})^a+b  x  (x^b–c)^b+c  x  (x^{c– a})^c+a

= (x ^{a²– b²} ) x     (x ^{b²– c²} ) x   (x ^{c²– a²})=x⁰=1. So, option (a) is correct.

4. If a^x ¸ a² = 1, then value of x is:

(a)      ½

(b)     2

(c)      1

(d)     0.

a^x ¸ a² = 1 = a⁰. So, a^{x– 2} = a⁰, or, x – 2 = 0    or, x = 2. So, option (b) is correct.

5. If x = 81, then value of (x^1/4 – 1) (x^1/4  + 1) will be:

(a)    6

(b)    8

(c)    2

(d)    1.

(x^1/4 – 1) (x^1/4  + 1) = {(81)^1/4 – 1} X {(81)^1/4 + 1}= {(3⁴)^1/4 – 1} X {(3⁴)^1/4 + 1} = (3 – 1) X ( 3 + 1) = 2 x 4 = 8. So, option (b) is correct.

6. If 3^{x– 1} = 81, then x will be:

(a)    8

(b)    5

(c)    3

(d)    2.

3^{x– 1} = 81, So, 3^{x– 1} = 81 = 3⁴. So, x – 1 = 4, Hence x = 4 + 1 = 5. So, option (b) is correct.

7. If 4^2x – 2^2x = 12 then x is:

(a)    6

(b)    1

(c)    5

(d)    3.

4^2x – 2^2x = 12, Or, (4^x)² – (2²)^x  = 12, Or, (4^x)² – (4)^x  = 12, Or, (4^x)² – 4^x – 12 = 0

Or, (4^x)² + 3X4^x – 4X 4^x – 12 = 0, Or, 4^x (4^x + 3) – 4 (4^x + 3) = 0, Or, (4^x + 3) (4^x – 4) = 0

Or, 4^x – 4 = 0, Or, 4^x = 4 = 4¹   So, x = 1. So, option (b) is correct.

8.Find value of (X^m+3n . x^4m-9n) / x^6m-6n

(a)    x^m

(b)    x^{– m}

(c)    xⁿ

(d)    1.

(x^m+3n . x^4m-9n) / x^6m-6n = (x^{m+ 3 n + 4 m – 9n}) / x^6m-6n  = x^{5m – 6n} / x^{6m – 6n} = x ^{5m – 6n – (6m – 6n)}

= x ^{5m – 6n – 6m + 6n} = x^-m.  So, option (b) is correct.

9. Value of  {x^{a(b – c)}  . x^{c(a – b)}} / x^{b(a– c)} is

(a)    a

(b)    1

(c)    x^a+b+c

(d)    0.

{x^{a(b – c)}  . x^{c(a – b)}}  / x^{b(a– c)} = x ^{(ab-ac)+(ac-bc) – (ab-ac)} = x ^{ab-ac+ac-bc – ab+bc} =x⁰=1. So, option (b) is correct.

10. Value of 16x ^{– 3} y² .  8^–1 _. x³ y ^–2 is

(a)      2xy

(b)     xy/2

(c)      2

(d)     1.

16x^{– 3} y² .  8^–1 _. x³ y ^–2 = 16 . 8^–1 . x^{– 3}  _. x³  . y² . y ^{– 2} = 16/8 . x⁰. y⁰ = 2 X 1 X 1=2. So, option (c) is correct.

11. Find the simplest value of 4 x 8 ^–2/3

a. 1

b. -1

c. ¼

d. ½

4 x 8 ^–2/3 = 2² X  ((2)³)^-2/3  = 2² X 2^(-2) = 2^2-2= 2⁰=1. So, option (a) is correct.

12. If 3^x = 2 ^{– x}, Find the value of x.

(a)    1

(b)    -1

(c)    0

(d)    6

3^x = 2 ^{– x}, or, 3^x = 1/2^x, Or, 3^x . 2^x = 1  Or,  (3 x 2)^x = 1  Or,  6^x = 1 = 6⁰  Or, x = 0. So, option (c) is correct.

13. If x = 8, y =27, find the value of (x^4/3  + y ^2/3)^1/2.

(a)    1

(b)    2

(c)    4

(d)    5.

Putting the value of x & y, we get (x^4/3  + y ^2/3)^1/2 = (8^4/3  + 27 ^2/3)^1/2 = {(2³) ^4/3 + (3³) ^2/3}^1/2 = (2⁴+3²)^1/2 = (16+9)^1/2 =(25)^1/2 = 5. So, option (d) is correct.

14. Simplify [(x^{a +  b–c}. x^{a – b + c}) ^b] ^c.

(a)    1

(b)    2

(c)    0

(d)    x^2abc.

[(x^{a +  b  –  c} . x^{a – b + c}) ^b] ^c = [(x^{a +  b –  c +  a – b + c}) ^b] ^c = {(x^2a) ^b]} ^c  = x^2abc. So, option (d) is correct.

15. Simplify (2^{x +1} + 2^x) / (2^{x +3} – 2^x+1)

_{(a)     2}

(b)     ½

(c)      1

(d)     0.

(2^{x +1} + 2^x) / (2^{x +3} – 2^x+1) = (2^x .  2¹ + 2^x) / (2^x . 2³ – 2^x . 2¹) = 2^x(2+1)/ 2^x (2³-2)

= 3 X 2^x / 6 X 2^x=3/6=1/2. So, option (b) is correct.

16. Simplify (x ^m+3n. x ^{4m – 9n}) / (x ^{6m – 6n})

(a)    X ^m

(b)    X ^–m

(c)    X ⁿ

(d)    X ^–n

(x ^m+3n. x ^{4m – 9n}) / (x ^{6m – 6n}) = x ^{(m+3n) + (4m-9n) – (6m-6n)} = x^(-m) . So, option (b) is correct.

17. Simplify 1/(1+z^a–b + z ^a–c) + 1/(1+z^b–c + z ^b–a )  + (1/1+z^c–a + z ^c–b)   

 (a)     1/3

(b)     1/2

(c)      1

(d)     0

  1/(1+z^a–b + z ^a–c) + 1/(1+z^b–c + z ^b–a ) + (1/1+z^c–a + z ^c–b)

= [z^{– a} / {z^{– a}(1+z^a–b + z ^a–c)}] + [z^-b / {z^-b (1+z^b–c + z ^b–a )}] + [z^-c / z^-c {(1 + z ^{c – a} + z ^{c – b} )}]

= [z^{– a} / (z^{– a} + z^{– b} + z^{– c} )] +  [z^{– b} / (z^{– b} + z^{– c} + z^{– a} )] +  [z^{– c} /( z^{– c} + z^{– a} + z^{– b})]

 = (z^{– a} + z^{– b} + z^{– c}) / (z^{– a} + z^{– b} + z^{– c}) =1. So, option (c) is correct

18. If 4^x = 5^y = 20^z then z =

(a)    20xy (b)    x+y/xy (c)    4/xy (d)    xy/x+y

Let 4x = 5y = 20z = k. Hence, 4 = k /x, 5 = k/y, 20 =  k/z

Or, 4 x 5 =  k/z, Or, k/x .  k/y =  k/z, Or,  k/x +k/y =  k/z. Or, k(1/x + 1/y) = k/z

Or, (1/x + 1/y) = 1/z or, (x+y)/xy = 1/z, or z= xy/(x+y). So, option (d) is correct

19. Find value of (3^½ / 9)^5/2 .  (9/ (3X 3 ^½ ) ^7/2 .  9   

(a)    1

(b)    4

(c)    9

(d)    27

(3^½ / 9)^5/2 .  (9/ (3X 3 ^½ ) ^7/2 .  9  = {3 ^½ . 3 ^{– 2}} ^5/2 . {3². 3 ^{– 3/2}} ^7/2 . {3²}

= (3^{– 2}) ^5/2. (3 ^½ )^7/2. 3² = 3 ^{– 15/4}. 3 ^7/4. 3² = 3 ^{– 15/4 + 7/4 +2} = 3 ^0/4 = 3⁰ =1. So, option (a) is correct

20. If 2^x – 2 ^x–1 = 4, then the value of x^x is

(a)    8

(b)    0

(c)    125

(d)    27

Let 2^x = a. So, 2^x – 2^{x -1}  = 4, a-(a/2)=4, or a/2=4. Or a=8. Now 2^x =8 = 2³. So, x=3. So, x^x=3 ³ = 27. So, option (d) is correct

21. If x = y ^a, y = z ^b and z = x ^c then abc =

(a)    2

(b)    1

(c)    0

(d)    5

x = y ^a = (z ^b)^a = {(x ^c)^b}^a = x^abc. So, x= x^abc. So, abc=1. So, option (b) is correct