This lesson will discuss two ways to solve logarithmic equations. The method you use depends on the type of logarithmic equation you are trying to solve.
The first type you will learn to solve has the following format
log_{b} X = Y
To solve it, you just have to write it in exponential form as X = b^{Y}Example #1
log (2x + 1) = 3
log_{10} (2x + 1) = 3
After writing it in exponential form we get
2x + 1 = 10^{3}
2x + 1 = 1000
2x = 999
x = 499.5
Check x = 499.5
log_{10} (2 x 499.5 + 1) = log_{10} (1000) = 3 since 10^{3} = 1000
Example #2
log_{5} (30x - 10) - 2 = log_{5} (x + 6)
You can solve this one the same way you solved example #1.
Start by putting all logarithms on one side of the equation.
log_{5} (30x - 10) - log_{5} (x + 6) - 2 = log_{5} (x + 6) - log_{5} (x + 6)
log_{5} (30x - 10) - log_{5} (x + 6) -2 = 0
log_{5} (30x - 10) - log_{5} (x + 6) -2 + 2 = 0 + 2
log_{5} (30x - 10) - log_{5} (x + 6) = 2
log_{5} [ (30x - 10) / (x + 6) ] = 2
Now that it has the format that we want, we can just write it in exponential form.
30x - 10 / x + 6 = 5^{2}
30x - 10 / x + 6 = 25 ( After multiplying both sides by x + 6 )
30x - 10 = 25 (x + 6)
30x - 10 = 25x + 150
30x - 25x = 150 + 10
5x = 160
x = 32
Check x = 32
log_{5} (30x - 10) - 2 = log_{5} (x + 6)
log_{5} (30 x 32 - 10) - 2 = log_{5} (32 + 6)
log_{5} (960 - 10) - 2 = log_{5} (32 + 6)
log_{5} (950) - 2 = log_{5} (38)
4.26 - 2 = 2.26
2.26 = 2.26
x = 32 is a good answer.
The second type you will learn to solve has the following format
log_{b} X = log_{b} Y
To solve it, you just have to see that log_{b} X = log_{b} Y if and only if X = Y
Example #3
log_{6} (2x - 4) + log_{6} (4) = log_{6} 40
log_{6} 4(2x - 4) = log_{6} 40
log_{6} 8x - 16 = log_{6} 40
8x - 16 = 40
8x = 40 + 16
8x = 56
x = 7
Check x = 7
log_{6} (2 x 7 - 4) + log_{6} (4) = log_{6} 40
log_{6} (14 - 4) + log_{6} (4) = log_{6} 40
log_{6} (10) + log_{6} (4) = log_{6} 40
log_{6} (10 x 4) = log_{6} 40
log_{6} (40) = log_{6} 40
x is 7 is a good answer.
Example #4
log_{3} x + log_{3} (x + 3) = log_{3} (2x + 6)
log_{3} x(x + 3) = log_{3} (2x + 6)
log_{3} (x^{2} + 3x) = log_{3} (2x + 6)
x^{2} + 3x = 2x + 6
x^{2} + 3x - 2x - 6 = 0
x^{2} + x - 6 = 0
(x - 2) (x + 3) = 0
x = 2 and x = -3
Check x = 2
log_{3} 2 + log_{3} (2 + 3) = log_{3} (2 x 2 + 6)
log_{3} 2 + log_{3} (5) = log_{3} (4 + 6)
log_{3} 2(5) = log_{3} (4 + 6)
log_{3} (10) = log_{3} (10)
x = 2 is a good answer.
Check x = -3
log_{3} -3 + log_{3} (-3 + 3) = log_{3} (2 x -3 + 6)
log_{3} -3 + log_{3} 0 = log_{3} 0
Since log_{3} -3 does not exist, x = -3 cannot be answer.
May 19, 19 09:20 AM
Basic math review game - The one and only math adventure game online. Embark on a quest to solve math problems!
Basic math formulas
Algebra word problems
New math lessons
Your email is safe with us. We will only use it to inform you about new math lessons.
Recommended
Scientific Notation Quiz
Graphing Slope Quiz
Adding and Subtracting Matrices Quiz
Factoring Trinomials Quiz
Solving Absolute Value Equations Quiz
Order of Operations Quiz
Types of angles quiz
May 19, 19 09:20 AM
Basic math review game - The one and only math adventure game online. Embark on a quest to solve math problems!