EXPONENTIAL FORM TO LOGARITHMIC FORM

Write the following equalities in logarithmic form.

Problem 1 :

8² = 64

Solution :

Given exponential form :

8² = 64

 x^y = k ó y = log_x k

Logarithmic form :

log₈ 64 = 2

Problem 2 :

10³ = 10000

Solution :

Given exponential form :

10³ = 10000

x^y = k ó y = log_x k

Logarithmic form :

 log₁₀ 10000 = 3

Problem 3 :

4^-2 = 1/16

Solution :

Given exponential form :

4^-2 = 1/16

x^y = k ó y = log_x k

Logarithmic form :

 log₄ (1/16) = -2

Problem 4 :

1/81 = 3^-4

Solution :

Given exponential form :

3^-4 = 1/81

x^y = k ó y = log_x k

Logarithmic form :

 log₃ (1/81) = -4

Problem 5 :

(1/2)^-5 = 32

Solution :

Given exponential form :

(1/2)^-5 = 32

x^y = k ó y = log_x k

Logarithmic form :

 log_1/2 32 = -5

Problem 6 :

(1/3)^-3 = 27

Solution :

Given exponential form :

(1/3)^-3 = 27

x^y = k ó y = log_x k

Logarithmic form :

 log_1/3 27 = -3

Problem 7 :

x^2x = y

Solution :

Given exponential form :

x^2z = y

x^y = k ó y = log_x k

Logarithmic form :

 log_x y = 2z

Problem 8 :

√x = y

Solution :

Given exponential form :

√x = y

Take square on both sides.

(x)^1/2 = y

x^y = k ó y = log_x k

Logarithmic form :

 log_x y = 1/2

Problem 9 :

Solve 5 ^{x - 2} = 1.7

Solution :

5 ^{x - 2} = 1.7

Taking logarithms on both sides, we get

log 5 ^{x - 2} = log 1.7

(x - 2) log 5 = log 1.7

x - 2 = log 1.7 / log 5

x - 2 = 0.2304/0.6989

x - 2 = 0.3296

x = 0.3296 + 2

x = 2.32

Approximately the value of x is 2.3.

Problem 10 :

A $1000 investment is made in a trust fund that pays 12% per annum, compounded monthly. How long will it take the investment to grow to $5000 ?

Solution :

Compound interest = Amount - interest

A = P(1 + i)^nt

Here i is rate of interest, n is number of times compounding and t is period of time investing.

A = 5000, P = 1000, i = 12/100 ==> 0.12

Converting for monthly = 0.12/12

= 0.01

n = 12, t = ?

5000 = 1000(1 + 0.01)^12t

Dividing by 1000, we get

5 = (1 + 0.01)^12t

log 5 = log (1.01)^12t

log 5 = 12t log (1.01)

log 5 / log 1.01 = 12t

0.6989/0.0043 = 12t

162.53 = 12t

t = 162.53/12

t = 13.5 years

Problem 11 :

Find the inverse of each of the following functions

a) f(x) = 2^x - 3

b) f(x) = 2⋅3^3x - 1

c) f(x) = -5⋅e^-x + 2

d) f(x) = 1 - 2 e^-2x

Solution :

a) f(x) = 2^x - 3

Let y = 2^x - 3

y + 3 = 2^x

Converting into logarithmic form, we get

log₂(y + 3) = x

x = log₂(y + 3)

Change x as f^-1(x) and y as x.

f^-1(x) = log₂(x + 3)

b) f(x) = 2⋅3^3x - 1

Let y = 2⋅3^3x - 1

y + 1 = 2⋅3^3x

Converting into logarithmic form, we get

(y + 1)/2 = 3^3x

log₃[(y + 1)/2] = 3x

x = (1/3) log₃[(y + 1)/2]

Change x as f^-1(x) and y as x.

f^-1(x) = (1/3) log₃[(x + 1)/2]

c) f(x) = -5⋅e^-x + 2

Let y = -5⋅e^-x + 2

y - 2 = -5⋅e^-x

Converting into logarithmic form, we get

-(y - 2)/5 = e^-x

-(y - 2)/5 = 1/e^x

(2 - y)/5 = 1/e^x

e^x = 1/[(2 - y)/5]

e^x = 5/(2 - y)

x = ln [5/(2 - y)]

Change x as f^-1(x) and y as x.

f^-1(x) = ln [5/(2 - x)]

d) f(x) = 1 - 2 e^-2x

Let y = 1 - 2 e^-2x

2 e^-2x = 1 - y

e^-2x = (1 - y) / 2

1/e^2x = [(1 - y) / 2]

e^2x = 2/(1 - y)

Converting into logarithmic form, we get

2x = ln [2/(1 - y)]

x = (1/2) ln [2/(1 - y)]

Change x as f^-1(x) and y as x.

f^-1(x) = (1/2) ln [2/(1 - x)]

Problem 12 :

One hundred grams of radium are stored in a container. The amount R (in grams) of radium present after t years can be modeled by R = 100e^−0.00043t. After how many years will only 5 grams of radium be present?

Solution :

R = 100e^−0.00043t

Amount of radium present = 5 grams

5 = 100e^−0.00043t

5/100 = e^−0.00043t

0.05 = e^−0.00043t

0.05 = 1/e^0.00043t

e^0.00043t = 1/0.05

e^0.00043t = 20

0.00043t = ln 20

0.00043t = 2.99

t = 2.99/0.00043

t = 6966 years approximately

Problem 13 :

The lengthℓ(in centimeters) of a scalloped hammerhead shark can be modeled by the function

ℓ = 266 − 219e^−0.05t

where t is the age (in years) of the shark. How old is a shark that is 175 centimeters long?

Solution :

ℓ = 266 − 219e^−0.05t

219e^−0.05t = 266 - l

e^−0.05t = (266 - l)/219

1/e^0.05t = (266 - l)/219

219/(266 - l) = e^0.05t

When l = 175

219/(266 - 175) = e^0.05t

219/91 = e^0.05t

ln (219/91) = 0.05t

ln (2.40) = 0.05t

0.87 = 0.05t

t = 0.87/0.05

t = 17.56

Approximately 18 years old.