SOLVING LOGARITHMIC EQUATIONS WITH BASE

To solve logarithmic equation, first we should know how to convert the logarithmic form to exponential form.

Step 1 :

Move the base to the other side of the equal sign.

Step 2 :

Create the same base on both sides.

Step 3 :

Using the properties of exponents, if the bases are same on both sides of the equal sign, then we can equate the powers.

Find the value of y.

Problem 1 :

log₅ 25 = y

Solution :

In logarithmic form :

log₅ 25 = y

Converting into exponential form :

5^y = 25

Write 25 as base of 5.

5^{y =} 5²

Since bases are equal, we equate the powers.

y = 2

Problem 2 :

log₃1 = y

Solution :

In logarithmic form :

log₃ 1 = y

Converting into exponential form :

3^y = 1

3^y = 3⁰ (3⁰ = 1)

Since bases are equal, we equate the powers.

y = 0

Problem 3 :

log₁₆ 4 = y

Solution :

In logarithmic form :

log₁₆ 4 = y

Converting into exponential form :

16^y = 4

(2⁴)^y = 2²

2^4y = 2²

4y = 2

y = 2/4

y = 1/2

Problem 4 :

log₂ (1/8) = y

Solution :

In logarithmic form :

log₂ (1/8) = y

Converting into exponential form :

2^y = 1/8

 2^y = 2^-3 (2^-3 = 1/8)

y = -3

Problem 5 :

log₅1 = y

Solution :

In logarithmic form :

log₅ 1 = y

Converting into exponential form :

5^y = 1

5^y = 5⁰ (5⁰ = 1)

y = 0

Problem 6 :

log₂ 8 = y

Solution :

In logarithmic form :

log₂ 8 = y

Converting into exponential form :

2^y = 8

2^{y =} 2³ (2³ = 8)

y = 3

Problem 7 :

log₇ (1/7) = y

Solution :

log₇ (1/7) = y

Converting from logarithmic form to exponential form, we get

7^y = 1/7

 7^y = 7 ^-1 (7^-1 = 1/7)

y = -1

Problem 8 :

log₃ (1/9) = y

Solution :

log₃ (1/9) = y

Converting from logarithmic form to exponential form, we get

3^y = 1/9

 3^y = 3^-2 (3^-2 = 1/9)

y = -2

Problem 9 :

log_y 32 = 5

Solution :

log_y 32 = 5

Converting from logarithmic form to exponential form, we get

y⁵ = 32

y⁵ = 2⁵

Since powers are equal, we equate the bases.

So,

y = 2

Problem 10 :

log₉ y = -1/2

Solution :

log₉ y = -1/2

9^-1/2 = y

(3²)^-1/2 = y

3^-1 = y

y = 1/3

Problem 11 :

log₄ (1/8) = y

Solution :

y = log_b x ó b^y = x

log₄ (1/8) = y

4^y = 1/8

(2²)^y = 1/2³

2^2y = 2^-3

2y = -3

y = -3/2

Problem 12 :

Log₉ (1/81) = y

Solution :

log₉ (1/81) = y

9^y = 1/81

 9^y = 9^-2

Since bases are equal, we equate the powers.

So,

y = -2

Problem 13 :

Describe the similarities and difference between in solving the equations 

4^{5x - 2} = 16 and log₄(10x + 6) = 1

Then solve the each equation

Solution :

4^{5x - 2} = 16 and log₄(10x + 6) = 1

In general by finding the inverse of exponential function, we will get the logarithmic function. 

Solving the exponential function :

4^{5x - 2} = 16

Writing 16 in exponential form, we get

4^{5x - 2} = 4²

5x - 2 = 2

5x = 2 + 2

5x = 4

x = 4/5

Solving logarithmic function :

log₄(10x + 6) = 1

10x + 6 = 4¹

10x + 6 = 4

10x = 4 - 6

10x = -2

x = -2/10

x = -1/5

Problem 14 :

For a sound with intensity I (in watts per square meter) the loudness L(I) of the sound (in decibels) is given by the function

L(I) = 10 log (I/I₀)

Where I0 is the intensity of barely audible sound (about 10^-12 watts per square meter) An artist in a recording studio turns up the volume of a track so that the intensity of the sound doubles. By how many decibels does the loudness increase ?

Solution :

L(I) = 10 log (I/I₀)

Let I be the original intensity then 2I is the double intensity

Increase in loudness = L(2I) - L(I)

= 10 log (2I/I₀) - 10 log (I/I₀)

Factoring 10, we get

= 10 [log (2I/I₀) - log (I/I₀)]

= 10 [log (2I) - log I₀ - (log I - log I₀)]

= 10 [log (2I) - log I₀ - log I + log I₀]

= 10 [log (2I) - log I]

= 10 [log 2 + log I - log I]

= 10 [log 2]

The loudness increases by 10 log 2 decibels or about 3 decibels.

Problem 15 :

The length ℓ (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

Solving for t, we get

219 e^−0.05t  = 266 - l

e^−0.05t  = (266 - l)/219

1/e^0.05t = [(266 - l)/219]

When l = 175 cm

219/(266 - l) = e^0.05t

e^0.05t  = 219/(266 - 175)

= 219/91

e^0.05t = 2.4

0.05t = ln (2.4)

0.05t = 0.87

t = 0.87/0.05

t = 17.5

Approximately 18 years.