SOLVING PROBLEMS ON LOGARITHMIC EQUATIONS

To solve a logarithmic equation, first combine the logarithmic terms using rules of logarithms.

Some of the rules in logarithm :

log m + log n = log (m x n)

log m - log n = log (m / n)

log mⁿ = n log m

log _a a = 1

After combining more than one logarithmic terms as one term, we have to convert logarithmic form to exponential form then solve for the variable.

Conversion from logarithmic form to exponential form :

Solve the following logarithmic equations.

Problem 1 :

2 logx = log 2 + log (3x - 4)

Solution :

2 logx = log 2 + log (3x - 4)

Let us combine the two terms on the right side

 logx² = log [2 ⋅ (3x - 4)]

x² = 6x - 8

x² - 6x + 8 = 0

x² - 4x - 2x + 8 = 0

x(x - 4) - 2(x - 4) = 0

(x - 4) (x - 2) = 0

x = 4 or x = 2

Hence the value of x is {2, 4}.

Problem 2 :

log x + log(x - 1) = log(4x)

Solution :

log x + log(x - 1) = log(4x)

log x + log(x - 1) = log 4 + log x

log(x - 1) = log 4

x - 1 = 4

x = 4 + 1

x = 5

Hence the value of x is 5.

Problem 3 :

log₃ (x + 25) - log₃ (x - 1) = 3

Solution :

log₃ (x + 25) - log₃ (x - 1) = 3

log a - log b = log (a/b)

log₃ [(x + 25)/(x -1)] = 3

[(x + 25)/(x -1)] = 3³

[(x + 25)/(x -1)] = 27

x + 25 = 27(x - 1)

x + 25 = 27x - 27

x - 27x = -27 - 25

-26x = -52

x = -52/(-26)

x = 2

Hence the value of x is 2.

Problem 4 :

log₉ (x - 5) + log₉ (x + 3) = 1

Solution :

log₉ (x - 5) + log₉ (x + 3) = 1

Let us combine the two terms on the left side

log₉ [(x - 5) (x + 3)] = 1

(x - 5) (x + 3) = 9

x² - 5x + 3x - 15 = 9

x² - 2x - 15 - 9 = 0

x² - 2x - 24 = 0

x² - 6x + 4x - 24 = 0

x(x - 6) + 4(x - 6) = 0

(x - 6) (x + 4) = 0

x = 6 or x = -4

Hence the value of x is 6 or -4.

Problem 5 :

log x + log (x - 3) = 1

Solution :

log x + log (x - 3) = 1

Let us combine the two terms on the left side

log [x (x - 3)] = 1

log (x² - 3x) = 1

x² - 3x = 10

x² - 3x - 10 = 0

x² - 5x + 2x - 10 = 0

x(x - 5) + 2(x - 5) = 0

(x - 5) (x + 2) = 0

x = 5 or x = -2

Hence the value of x is 5.

Problem 6 :

log₂ (x - 2) + log₂ (x + 1) = 2

Solution :

log₂ (x - 2) + log₂ (x + 1) = 2

Let us combine the two terms on the left side

log₂ [(x - 2) (x + 1)] = 2

log_a b = x ó b = a^x

(x - 2) (x + 1) = 2²

(x - 2) (x + 1) = 4

x² - 2x + x - 2 = 4

x² - x - 2 - 4 = 0

x² - x - 6 = 0

x² - 3x + 2x - 6 = 0

x(x - 3) + 2(x - 3) = 0

(x + 2) (x - 3) = 0

x = -2 or x = 3

x = 3

Hence the value of x is 3 or -2.

Problem 7 :

ln x = -3

Solution :

ln x = -3

x = e^-3

Problem 8 :

log(3x - 2) = 2

Solution :

log₁₀(3x - 2) = 2

log_mx = y ó x = m^y

3x - 2 = 10²

3x - 2 = 100

3x = 100 + 2

3x = 102

x = 102/3

x = 34

Hence the value of x is 34.

Problem 9 :

log (3x - 2) = 2

Solution :

log₁₀(3x - 2) = 2

3x - 2 = 10²

3x - 2 = 100

3x = 100+2

3x = 102

x = 102/3

x = 34

Problem 10 :

log₅(x + 6) + log₅(x + 2) = 1

Solution :

log₅(x + 6) + log₅(x + 2) = 1

log₅(x + 6)(x + 2) = 1

x² + 2x + 6x +12 = 5¹

x² + 8x + 12 - 5 = 0

x² + 8x + 7 = 0

(x + 1(x + 7) = 0

x = -1, x = -7