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In this lesson, we will look into how to solve logarithmic equations using the Properties of Logarithms.

**Properties of Logarithms - Part 1**
**Properties of Logarithms - Part 2 - Solving Logarithmic Equations**

More Lessons for Algebra

Math Worksheets

In this lesson, we will look into how to solve logarithmic equations using the Properties of Logarithms.

An equation that contains a logarithm of a variable quantity is called a logarithmic equation. Logarithmic equations can generally be solved using the properties of logarithm and the following property.

For two logarithms of the same base,

Log

log_{a}M =↔_{a}NM = N

** Example: **

Solve the logarithmic equation

log _{2} (*x* – 1) + log _{2} (*x* – 4) = log _{2} (2*x* – 6)

** Solution: **

log _{2} (*x* – 1) + log _{2} (*x* – 4) = log _{2} (2*x* – 6)

log _{2} (*x* – 1)(*x* – 4) = log _{2} (2*x* – 6)

(*x* – 1)(*x* – 4) = (2*x* – 6)

*x*^{2} – 5*x* + 4 = 2*x* – 6

*x*^{2} – 7 *x* + 10 = 0

(*x* – 2)(*x* – 5) = 0

*x* = 2 or 5

We need to check whether each logarithm is defined for these values of *x*.

When *x* = 2,

log _{2} (*x* – 1) = log _{2 }1

log _{2} (*x* – 4) = log _{2 }(– 2) , whuch is undefined

log _{2} (2*x* – 6) = log _{2} _{ }(– 2) , whuch is undefined

So, *x* = 2 is rejected

When *x* = 5,

log _{2} (*x* – 1) = log _{2 }4

log _{2} (*x* – 4) = log _{2 }1

log _{2} (2*x* – 6) = log _{2} _{ }4

So, *x* = 5 is the answer

**Example:**

Solve the logarithmic equation

log _{3} 2 + log _{3} (*x* + 4) = 2log _{3} *x *

**Solution:**

log _{3} 2 + log _{3} (*x* + 4) = 2log _{3} *x
*log

(

*x* = –2 is rejected because log _{3} (–2) is undefined

So the answer is *x* = 4

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