Sign up

Login
Home

Math

Physics

Chemistry

Biology

Other

Tools
Submit New Formulas/Articles
Recently Added Math Formulas
·
Conic Section
·
Exponential Functions
·
Logarithmic Identities
·
Polynomials Second Degree
·
Polynomials Basic
Additional Formulas
·
Closure Property Addition
·
Closure Property  Multiplication
·
Properties of Equality
·
Powers
·
Polynomials Basic
·
Polynomials Second Degree
·
Logarithmic Identities
·
Exponential Functions
·
Conic Section
Current Location
>
Math Formulas
>
Algebra
> Logarithmic Identities
Logarithmic Identities
log (10) = 1
log (1) = 0
log(ab) = log(a) + log(b)
log(a/b) = log(a)  log(b)
log(am) = m ∙ log(a)
log(m√(a)) = 1/m ∙ log(a)
For natural logarithm (ln) functions the following identities apply
ln (10) = 1
ln (1) = 0
ln(ab) = ln(a) + ln(b)
ln(a/b) = ln(a)  ln(b)
ln(am) = m ∙ ln(a)
ln(m√(a)) = 1/m ∙ ln(a)
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.
WebFormulas.com ©
2024

Contact us

Terms of Use

Privacy Policy
