IB Math SL Exponents and Logarithms Questions

Question 1 :

Simplify :

(16y⁴/x^-8)^1/4

Answer :

= (16y⁴/x^-8)^1/4

= (16x⁸y⁴)^1/4

= (16)^1/4(x⁸)^1/4(y⁴)^1/4

= (2⁴)^1/4(x⁸)^1/4(y⁸)^1/4

= 3xy²

Question 2 :

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

Answer :

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

Simplify.

= (2 ⋅ 3)(y^1/2 ⋅ y^-1)

= 6y^{1/2 - 1}

= 6y^-1/2

= 6/y^1/2

Substitute y = 4.

= 6/(4)^1/2

= 6/(2²)^1/2

= 6/2

= 3

Question 3 :

Simplify :

(x³y²)^-a(x^ay^-b)³

Answer :

= (x³y²)^-a(x^ay^-b)³

= (x³)^-a(y²)^-a(x^a)³(y^-b)³

= x^-3ay^-2ax^3ay^-3b

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

= x⁰y^{-(2a + 3b)}

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

Question 4 :

Solve for x :

3^{x + 6} = 27^x

Answer :

3^{x + 6} = (3³)^x

Using power of the power rule,

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

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 5 :

Solve for x :

(√25)^-7(√5)^-5 = 5^k

Answer :

(√25)^-7(√5)^-5 = 5^k

5^-7(5^1/2)^-5 = 5^k

Using power of the power rule,

5^-75^-5/2 = 5^k

Using product rule of derivative,

5^{-7 + (-5/2)} = 5^k

5^{-7 - 5/2} = 5^k

5^-19/2 = 5^k

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

k  = -19/2

Question 6 :

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

Question 7 :

Evaluate :

log₂64

Answer :

log₂64 = log₂(2⁶)

Using the power rule of logarithms,

= 6log₂2

= 6(1)

= 6

Question 8

Evaluate :

log_3√3729

Answer :

Write 729 as a power of 3√3.

729 = 3⁶

= 3^{4 + 2}

= 3⁴ ⋅ 3²

= 3⁴ ⋅ [(√3)²]²

= 3⁴ ⋅ (√3)⁴

= (3√3)⁴

log_3√3729 = log_3√3(3√3)⁴

= 4log_3√3(3√3)

= 4(1)

= 4

Question 9

Given that

log₁₀2 = x

log₁₀3 = y

Write log₁₀60 in terms of x and y.

Answer :

log₁₀60 = log₁₀(2 ⋅ 3 ⋅ 10)

Using the product rule of logarithm,

= log₁₀2 + log₁₀3 + log₁₀10

= log₁₀2 + log₁₀3 + 1

Substitute log₁₀2 = x and log₁₀3 = y.

= x + y + 1

Question 10

Given that

log₁₀2 = x

log₁₀3 = y

Write log₁₀1.2 in terms of x and y.

Answer :

log₁₀1.2 = log₁₀(12/10)

Using the quotient rule of logarithm,

= log₁₀12 + log₁₀10

= log₁₀(2 ⋅ 2 ⋅ 3) + 1

Using the product rule of logarithm,

= log₁₀2 + log₁₀2 + log₁₀3 + 1

Substitute log₁₀2 = x and log₁₀3 = y.

= x + x + y + 1

= 2x + y + 1

Question 11 :

If 2log₁₀x = 4log₁₀3, then find the value of x.

Answer :

2log₁₀x = 4log₁₀3

Divide each side by 2.

log₁₀x = 2log₁₀3

Using the power rule of logarithm,

log₁₀x = log₁₀3²

log₁₀x = log₁₀9

The above two logarithms are equal with the same base. So, the arguments can be equated.

x = 9

Question 12 :

Find the value of x, if log_x5 = 1/2.

Answer :

log_x5 = 1/2

The above equation is in logarithmic form. Convert it to exponential form to solve for d.

5 = x^1/2

Raise the exponent to 2 on both sides.

(x^1/2)² = 5²

Using power of a power rule,

x¹ = 25

x = 125

Question 13 :

Find f^-1(x), if f(x) = -2lln(5 - 2x) + 8.

Answer :

f(x) = -2lln(5 - 2x) + 8

Replace f(x) by y.

y = -2lln(5 - 2x) + 8

Interchange x and y. 

x = -2lln(5 - 2y) + 8

Subtract 8 from both sides.

x - 8 = -2ln(5 - 2y)

Divide both sides by -2.

(x - 8)/(-2) = ln(5 - 2y)

(8 - x)/2 = ln(5 - 2y)

In the equation above, we have natural logarithm 'ln' and its base is e. Convert it to exponential form.

e^{(8 - x)/2} = 5 - 2y

Add 2y to both sides.

e^{(8 - x)/2} + 2y = 5

Subtract e^{(8 - x)/2} from both sides.

2y = 5 - e^{(8 - x)/2}

Divide both sides by 5.

y = (5 - e^{(8 - x)/2})/2

Replace y by f^-1(x).

f^-1(x) = (5 - e^{(8 - x)/2})/2

Question 14 :

It is thought that a computer virus would spread in an office according to an exponential model. 6 computers were infected initially. The number of computers infected hearing it grows at a rate of 11% per hour.

(a) How many computers were infected after 24 hours? 

There are 8,200 computers in the office.

(b) How long it would take for all computers in the office to be infected.

Answer :

Part (a) :

From the given information, we can write the function which gives the number of computer infected after t hours.

I(t) = 6(1 + 0.11)^t

I(t) = 6(1.11)^t

Substitute t = 24.

I(24) = 6(1.11)²⁴

I(24) ≈ 73

After 24 hours, about 73 computers were infected.

Part (b) :

I(t) = 6(1.11)^t

6(1.11)^t = I(t)

Substitute I(t) = 8200.

6(1.11)^t = 8200

Divide both sides by 6.

1.11^t = 4100/3

Take logarithm on both sides.

log(1.11^t) = log(4100/3)

tlog1.11 = log4100 - log3

t(0.0453) = 3.6128 - 0.4771

0.0453t = 3.1357

Divide both sides by 0.0453.

t ≈ 69

It would take about 69 hours for all computers in the office to be infected.

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