IB DIPLOMA MATHEMATICS - Problems on Exponents

Problem 1 to 5 : Simplify the given expression.

Problem 1 :

12x² ⋅ 4x⁵

Solution :

= 12x² ⋅ 4x⁵

= (12 ⋅ 4)(x² ⋅ x⁵)

= 48x^{2 + 5}

= 48x⁷

Problem 2 :

Solution :

Problem 3 :

Solution :

Problem 4 :

Solution :

= x(2 - 7x⁵)

Problem 5 :

Solution :

= 4ac(5ab⁵ - 4c)

Problem 6 to 10 : Simplify the given expression.

Problem 6 :

4^{2x - 7} = 16

Solution :

4^{2x - 7} = 16

4^{2x - 7} = 4²

Since the bases is same on both siodes, exponents can be equated.

2x - 7 = 2

Add 7 to both sides.

2x = 9

Divide both sides by 2.

x = 4.5

Problem 7 :

Solution :

3^{2x + 5} = 3^-3

Since the bases is same on both siodes, exponents can be equated.

2x + 5 = -3

Subtract 5 from both sides.

2x = -8

Divide both sides by 2.

x = -4

Problem 8 :

10 + 2 ⋅ 2^x = 18

Solution :

10 + 2 ⋅ 2^x = 18

10 + 2^{1 + x} = 18

Subtract 10 from both sides.

2^{1 + x} = 8

2^{1 + x} = 2³

Since the bases is same on both siodes, exponents can be equated.

1 + x = 3

Subtract 1 from both sides.

x = 2

Problem 9 :

6^x = 81 ⋅ 2^x

Solution :

6^x = 81 ⋅ 2^x

(2 ⋅ 3)^x = 3⁴ ⋅ 2^x

2^x ⋅ 3^x = 3⁴ ⋅ 2^x

Divide both sides by 2^x.

3^x = 3⁴

Since the bases is same on both siodes, exponents can be equated.

x = 4

Problem 10 :

32 + 2^{x - 1} = 2^x

Solution :

32 + 2^{x - 1} = 2^x

64 + 2^x = 2 ⋅ 2^x

Let y = 2^x.

64 + y = 2y

Subtract y from both sides.

64 = y

Substitute 2^x for y.

64 = 2^x

2⁶ = 2^x

Since the bases is same on both siodes, exponents can be equated.

x = 6

Problem 11 :

Solve the simultaneous equations :

8^x2^y = 1

Solution :

(1) + (2) :

5x = 5

x = 1

Substitute x = 1 into (1).

3(1) + y = 0

3 + y = 0

y = -3

Therefore, the solution is

(x , y) = (1, -3)

Problem 12 :

The length of a baby fish is modelled by L = 2t² where t is the age in days and L is the length in cm. Its mass in grams is modelled by M = 4L³.

(A) Find and simplify an expression for M in terms of t.

(B) Find the age of the fish when the model predicts a mass of 1000 g.

(C) Explain why the model is unlikely to still hold after 100 days.

Solution :

Part A :

M = 4L³

Substitute 2t² for L.

M = 4(2t²)³

M = 4(2)³(t²)³

M = 4 ⋅ 8 ⋅ t⁶

M = 32t⁶

Part B :

M = 32t⁶

Substitute 1000 for M and solve for t..

32t⁶ = 1000

t ≈ 1.77 days

Part C :

The model predicts that the fish will continue growing, whereas in reality it is likely that after 100 days the fish
will be growing far more slowly, if at all.

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