IB Math AA SL Questions on Exponents

Question 1 :

Simplify :

(81x⁴/y^-8)^1/4

Answer :

= (81x⁴/y^-8)^1/4

= (81x⁴y⁸)^1/4

= (81)^1/4(x⁴)^1/4(y⁸)^1/4

= (3⁴)^1/4(x⁴)^1/4(y⁸)^1/4

= 3xy²

Question 2 :

Evaluate (2x^1/2)(3x^-1), if x = 4.

Answer :

= (2x^1/2)(3x^-1)

Simplify.

= (2 ⋅ 3)(x^1/2 ⋅ x^-1)

= 6x^{1/2 - 1}

= 6x^-1/2

= 6/x^1/2

Substitute x = 4.

= 6/(4)^1/2

= 6/(2²)^1/2

= 6/2

= 3

Question 3 :

Simplify :

(6ab²c³)(4b^-2c^-3d)

Answer :

= (6ab²c³)(4b^-2c^-3d)

= (6 ⋅ 4)(a)(b² ⋅ b^-2)(c³ ⋅ c^-3)(d)

= 24ab^2-2c^3-3d

= 24ab⁰c⁰d

= 24a(1)(1)d

= 24ad

Question 4 :

Simplify :

(x^ay^-b)³(x³y²)^-a

Answer :

= (x^ay^-b)³(x³y²)^-a

= (x^a)³(y^-b)³(x³)^-a(y²)^-a

= x^3ay^-3bx^-3ay^-2a

= x^{3a - 3a}y^{-2a - 3b}

= x⁰y^{-(2a + 3b)}

= 1/y^{2a + 3b}

Question 5 :

Solve for x :

2^{x + 6} = 8^x

Answer :

2^{x + 6} = (2³)^x

Using power of the power rule,

2^{x + 6} = 2^3x

We have the same base on both sides. So, the exponents can be equated.

x + 6 = 3x

Subtract 3x from both sides.

-2x + 6 = 0

Subtract 6 from both sides.

-2x = -6

Divide both sides by -2.

x = 3

Question 6 :

Solve for x :

27^{5 - 3x} = 1/[(√(3^{6x - 2})]

Answer :

27^{5 - 3x} = 1/[(√(3^{6x - 2})]

(3³)^{5 - 3x} = 1/(3^{6x - 2})^1/2

Using power of the power rule,

3^{15 - 9x} = 1/(3^{3x - 1})

3^{15 - 9x} = 3^{-(3x - 1)}

3^{15 - 9x} = 3^{-3x + 1}

We have the same base on both sides. So, the exponents can be equated.

15 - 9x = -3x + 1

Add 3x to both sides.

15 - 6x = 1

Subtract 15 from both sides.

-6x = -14

Divide both sides by -6.

x = 7/3

Question 7 :

Solve for x :

x√x = (x√x)^x

Answer :

x√x = (x√x)^x

For any value, if there is no exponent, the exponent is 1. For example, consider y. Here, there is no exponent for y. So, the exponent of y can be taken as 1.

On the left side of x√x = (x√x)^x, we can take the exponent 1 for x√x.

(x√x)¹ = (x√x)^x

We have the same base on both sides. So, the exponents can be equated.

x = 1

Question 8 :

The length of a snake 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 snake when the model predicts a mass of 1000 g.

Answer :

Part (a) :

M = 4L³

Substitute L = 2t².

M = 4(2t²)³

M = 4(2³)(t²)³

M = 4(8)(t⁶)

M = 32t⁶

Part (b) :

M = 32t⁶

Substitute M = 1000 and solve for t.

1000 = 32t⁶

Divide both sides by 32.

t⁶ = 1000/32

Take sixth root on both sides.

t = ⁶√(1000/32)

t ≈ 1.77 days

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