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.
Nov 15, 18 05:01 PM
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Nov 15, 18 05:01 PM
Modeling multiplication with number counters - Learning multiplication is fun!
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