Solving Logarithmic Equations

A logarithm is defined as the exponent to which a given base must be raised to obtain a specific number. It represents an equivalent form of an exponential expression and is written as:

\log_b a = x

where b is the base, a is the argument, and x is the logarithm, with b > 0, b ≠ 1, and a > 0.

Some methods for solving logarithmic equations are the following:

1. Converting to Exponential Form

One of the most effective methods to solve logarithmic equations is to convert them into exponential form. The logarithmic equation:

\log_b(x) = y

This can be rewritten in its exponential form:

x = b^y

This transformation allows us to work with the equation in a different but equivalent format, often making it easier to solve.

Example: Solve,\log_2(x) = 3.
Solution:

Convert to exponential form: x = 2³
Simplify: x = 8

2. Using Properties of Logarithms

Using the properties of logarithms, we can help break down more complex logarithmic equations into simpler forms.

Example: Solve log₃(9x) = 5.
Solution:

Use the product property: log⁡₃(9) + log₃(x) = 5

Simplify log₃(9): log₃(3²) = 2 · log₃(3) = 2 [log₃(3) = 1]

Substitute back into the equation:
2 + log⁡₃(x) = 5
log⁡₃(x) = 5 - 2 = 3

Convert to exponential form: x = 3³
x = 27

3. Isolating the Logarithmic Expression

To solve a logarithmic equation, it is often necessary to isolate the logarithmic expression first.

Example: Solve log⁡₅(x − 3) + 2 = 4.
Solution:

Isolate the logarithmic expression: log₅(x - 3) = 4 - 2 = 2
Convert to exponential form: x − 3 = 5²
Simplify: x − 3 = 25
Solve for x: x = 25 + 3 = 28

Related Articles

• Exponential Function
• Logarithmic Formulas
• Log Rules

Solved Examples

Example 1: Solve log₂ x (x) = 3
Solution:

Convert the logarithmic equation to its exponential form:

2³ = x
⇒ 8 = x

Therefore, the solution is: x = 8

Example 2: Solve log₃ (x) + log₃ (x - 2) = 1.
Solution:

Use the product property of logarithms: log_b(mn) = log_b(m) + log_b(n)
log₃(x(x - 2)) = 1

Convert the logarithmic equation to its exponential form:
3¹ = x(x - 2)

Simplify and solve the quadratic equation:
3 = x² - 2x
⇒ x² - 2x - 3 = 0

Factor the quadratic equation:
(x - 3)(x + 1) = 0

Solve for x:
x = 3 or x = -1

Since the logarithm of a negative number is undefined, the solution is:
x = 3

Example 3: Solve log₅(x + 1) + log₅(x - 1) = 1.
Solution:

Use the product property of logarithms:
log₅ ((x + 1)(x - 1)) = 1

Convert the logarithmic equation to its exponential form:
5¹ = (x + 1)(x - 1)

Simplify the right-hand side:
5 = x² - 1

Solve the quadratic equation:
x² - 1 = 5
⇒ x² = 6
x = ± √6

Since log₅ (x + 1) and log₅ (x - 1) must both be defined, x must be greater than 1. Therefore, the solution is:
x = √6

Example 4: Solve log₂(x + 3) - log_₂(x - 1) = 2
Solution:

Use the quotient property of logarithms: log_b (m/n) = log_b (m) - log_b (n)
\log_2 \left(\frac{x + 3}{x - 1}\right) = 2

Convert the logarithmic equation to its exponential form:
2² = \frac{x + 3}{x - 1}

Simplify and solve the resulting equation:

4 = \frac{x + 3}{x - 1}
⇒ 4(x - 1) = x + 3
⇒ 4x - 4 = x + 3
⇒ 3x = 7
⇒ x = 7/3

Example 5: Solve log₄ (2x - 1) + log₄ (x + 3) = 2
Solution:

Use the product property of logarithms:
log₄ ((2x - 1)(x + 3)) = 2

Convert the logarithmic equation to its exponential form:
4² = (2x - 1)(x + 3)

Simplify and solve the resulting quadratic equation:
16 = (2x - 1)(x + 3)
⇒ 16 = 2x² + 6x - x - 3
⇒ 16 = 2x² + 5x - 3
⇒ 2x² + 5x - 19 = 0

Solve the quadratic equation using the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Here, a = 2, b = 5, and c = -19:

x = \frac{-5 \pm \sqrt{25 + 152}}{4}

For logs to exist:

• 2x − 1 > 0 ⇒ x > 1/2
• x + 3 > 0 ⇒ x > −3

So overall: x > 1/2,

Reject the negative root ⇒ x = \frac{-5 + \sqrt{177}}{4}