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Solving logarithmic functions using Logarithmic Identities
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Solving logarithmic functions using Logarithmic Identities
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Solving logarithmic functions using Logarithmic Identities
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Example 1: Solve
log
_{2}
(x) + log
_{2}
(x – 2) = 3
Solution
: Here we need to use logarithmic identities to combine the two terms on the lefthand side of the equation:
log
_{2}
(x) + log
_{2}
(x – 2) = 3
log
_{2}
((x) (x – 2)) = 3
log
_{2}
(x
^{2}
– 2x) = 3
Then we’ll use the relationship to convert the log form to the corresponding exponential form, and then move further,
log
_{2}
(x
^{2}
– 2x) = 3
2
^{3}
=
x
^{2}
– 2
x
8 =
x
^{2}
– 2
x
0 =
x
^{2}
– 2
x
– 8
0 = (
x
– 4) (
x
+ 2)
x
= 4, –2
But if
x = –2
, then "
log
_{2}
(x)
", from the original logarithmic equation, will have a negative number for its argument (as will the term "
log
_{2}
(x – 2)
"). Since logs cannot have zero or negative arguments, then the solution to the original equation cannot be
x = –2
.
The solution is x
= 4
.
Example 2: Solve
log
_{2}
(x
^{2}
) = (log
_{2}
(x))
^{ }
^{2}
.
Solution
: First, we'll write out the square on the righthand side:
log
_{2}
(x
^{2}
) = (log
_{2}
(x))
^{ }
^{2}
log
_{2}
(x
^{2}
) = (log
_{2}
(x)) (log
_{2}
(x))
Then we’ll apply the log rule to move the "squared", from inside the log on the lefthand side of the equation, out in front of that log as a multiplier. Then we'll move that term to the righthand side:
2log
_{2}
(x) = [log
_{2}
(x)] [log
_{2}
(x)]
0 = [log
_{2}
(x)] [log
_{2}
(x)] – 2log2(x)
Factorize the above equation to obtain
0 = [log
_{2}
(x)] [log
_{2}
(x) – 2]
log
_{2}
(x) = 0 or log2(x) – 2 = 0
2
^{0}
= x or log2(x) = 2
1 = x or 22 = x
1 = x or 4 = x
Example 3: Solve
Solution:
Since
4/x
is the base of a log function x is greater than 0.
The equation is then equivalent with
(4/x)
^{ }
^{2}
= x
^{2}
 6
<=>
16 = x
^{4}
 6x
^{2}
^{ }
Now, let
x
^{2}
= t
<=>
16 = t
^{2}
 6t
<=>
t=2 or t=8
<=>
x = sqrt (8)
(since x >0)
The solution is
x = 1, 4.
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