1. log 3 x = 2
3 ^ (2) = x
x = 3 ^ (2)
x = 9
2. 10^(log(27)) = x
x = 10^(log27)
log x = log 10^(log27)
log x = log27 * log 10
log x = log27 * 1
x = 27
3. 3^(2x−1) = 27 ^ (2x−1)
log 3 ( 3 ^ (2x-1) ) = log 3 ( 27 ^ (2x-1) )
(2x-1) * log 3 3 = (2x-1) * log 3 27
(2x-1) * 1 = (2x-1) *3
2x-1 = 6x-3
2x-6x = -3+1
-4x = -2
x = 1/2
2015년 11월 3일 화요일
Logarithm Law 6
Logarithm Law 6
log a (x)
log a ( x ^ ( 1 / b ) ) = ---------
b
The derivation of this law is identical to the derivation of Logarithm Law 5
For example, we can show that log 2 ( 5√ 7) = log 2 (7) / 5.
log 2 ( 5√ 7) = log 2 ( 7 ^( 1/5 ) )
= 1/5 log 2 (7)
= log 2 (7) / 5
Therefore, log 2 ( 5√ 7) = log 2 (7) / 5
1. log 2 (4√ 8) = log 2 ( 8 ^ (1/4) )
= 1/4 log 2 (8)
= 1/4 log 2 ( (2)^3 )
= 3/4 log 2 (2)
= 3/4
2. log 8 (10√10) = log 8 (10 ^ (1/10))
= 1/10 log 8 (10)
3. log 16 (y√ x) = log 16 (x ^ (1/y))
= 1/y log 16 (x)
4. log z ( x√ y) = log z (y ^ (1/x))
= 1/x log z (y)
5. log x ( 2x √ y) = log x (y ^ (1/2x))
= 1/2x log x (y)
log a (x)
log a ( x ^ ( 1 / b ) ) = ---------
b
The derivation of this law is identical to the derivation of Logarithm Law 5
For example, we can show that log 2 ( 5√ 7) = log 2 (7) / 5.
log 2 ( 5√ 7) = log 2 ( 7 ^( 1/5 ) )
= 1/5 log 2 (7)
= log 2 (7) / 5
Therefore, log 2 ( 5√ 7) = log 2 (7) / 5
1. log 2 (4√ 8) = log 2 ( 8 ^ (1/4) )
= 1/4 log 2 (8)
= 1/4 log 2 ( (2)^3 )
= 3/4 log 2 (2)
= 3/4
2. log 8 (10√10) = log 8 (10 ^ (1/10))
= 1/10 log 8 (10)
3. log 16 (y√ x) = log 16 (x ^ (1/y))
= 1/y log 16 (x)
4. log z ( x√ y) = log z (y ^ (1/x))
= 1/x log z (y)
5. log x ( 2x √ y) = log x (y ^ (1/2x))
= 1/2x log x (y)
Logarithm law 5
Logarithm law 5
log a ( x ^ (b) ) = b log a (x)
we need to relate x to the base a. So, we let x = a ^ (n). Then,
log a ( x ^ (b) ) = log a ( ( (a) ^ n ) ^b )
= log a ( (a) ^ (n * b) )
log a ( x ^ (b) ) = b log a (x)
we need to relate x to the base a. So, we let x = a ^ (n). Then,
log a ( x ^ (b) ) = log a ( ( (a) ^ n ) ^b )
= log a ( (a) ^ (n * b) )
Assumption that x = a ^ (n)
n = log a (x)
Therefore
log a (x ^ (b) ) = log a ( a ^ ( b * log a (x) ) )
= b * log a (x)
For example, we can show that log 2 ( (3) ^ 5 ) = 5 log 2 (3).
log 2 ( (3) ^ 5 ) = log 2 (3*3*3*3*3)
= log 2 (3) + log 2 (3) + log 2 (3) + log 2 (3) + log 2 (3)
= 5 log 2 (3)
Therefore, log 2 ( (3) ^ 5 ) = 5 log 2 (3)
For example, we can show that log 2 ( (5) ^ 2 ) = 2 log 2 (5).
log 2 ( (5) ^ 2 ) = log 2 (5*5)
= log 2 (5) + log 2 (5)
= 2 log 2 (5)
Therefore, log 2 ( (5) ^ 2 ) = 2 log 2 (5)
Logarithm law 3
log a (x*y) = log a (x) + log a (y)
The derivation of this law is a bit trickier than the first two. Firstly, we need to relate x and y to the base a. So, assume that x = a ^ (m) and y = a ^ (n).
log a (x) = m
log a (y) = n
This means that we can write:
log a (x*y) = log a ( a^(m) . a^(n) )
= log a ( a ^ (m+n) )
= log a ( a ^ ( log a (x)+log a (y) )
= log a (x) + log a (y)
For example, show that
log ( 10 * 0.1 ) = log 10 + log 0.1
= log ( 1 )
= log ( 10 ^ (0) )
= 0
log ( 10 * 1 ) = log 10 + log 1
= log ( 10 )
= log ( 10 ^ (1) )
= 1
log ( 10 * 10 ) = log 10 + log 10
= log ( 100 )
= log ( 10 ^ (2) )
= 2
log ( 10 * 100 ) = log 10 + log 100
= log ( 1000 )
= log ( 10 ^ (3) )
= 3
The derivation of this law is a bit trickier than the first two. Firstly, we need to relate x and y to the base a. So, assume that x = a ^ (m) and y = a ^ (n).
log a (x) = m
log a (y) = n
This means that we can write:
log a (x*y) = log a ( a^(m) . a^(n) )
= log a ( a ^ (m+n) )
= log a ( a ^ ( log a (x)+log a (y) )
= log a (x) + log a (y)
For example, show that
log ( 10 * 0.1 ) = log 10 + log 0.1
= log ( 1 )
= log ( 10 ^ (0) )
= 0
log ( 10 * 1 ) = log 10 + log 1
= log ( 10 )
= log ( 10 ^ (1) )
= 1
log ( 10 * 10 ) = log 10 + log 10
= log ( 100 )
= log ( 10 ^ (2) )
= 2
log ( 10 * 100 ) = log 10 + log 100
= log ( 1000 )
= log ( 10 ^ (3) )
= 3
Natural Logarithm
The natural logarithm (symbol ln) is widely used in the sciences.
While the notation log10(x) and loge(x) may be used, log10(x) is often written log(x) in Science and loge(x) is normally written as ln(x) in both Science and Mathematics. So, if you see the log symbol without a base, it means log10.
Logarithms can be changed from one base to another, by using the change of base formula:
log y x
log a x = --------
log y a
where y is any base you find convenient.
Normally a and y are known, therefore logy a is normally a known, if irrational, number.
for example :
log 10 13
log 4 13 = --------
log 10 4
log 10 16
log 4 16 = --------
log 10 4
While the notation log10(x) and loge(x) may be used, log10(x) is often written log(x) in Science and loge(x) is normally written as ln(x) in both Science and Mathematics. So, if you see the log symbol without a base, it means log10.
Logarithms can be changed from one base to another, by using the change of base formula:
log y x
log a x = --------
log y a
where y is any base you find convenient.
Normally a and y are known, therefore logy a is normally a known, if irrational, number.
for example :
log 10 13
log 4 13 = --------
log 10 4
log 10 16
log 4 16 = --------
log 10 4
Logarithms introduction
Logarithms, commonly referred to as logs, are the inverse of exponentials.
The logarithm of a number x in the base a is defined as the number n such that a^(n) = x
So, if a^(n) = x, then :
log_a (x) = n
The logarithm of a number x in the base a is defined as the number n such that a^(n) = x
So, if a^(n) = x, then :
log_a (x) = n
Logarithm law 4
log a (x/y) = log a (x) - log a (y)
The derivation of this law The derivation of this law is identical to the derivation of Logarithm Law 3 and is left as an exercise.
For example, show that log ( 10/1000 ) = log 10 − log 1000.
log (10)
log (1000)
log (10 / 1000) = log (1 / 100)
= log (10^(-2))
= -2
The derivation of this law The derivation of this law is identical to the derivation of Logarithm Law 3 and is left as an exercise.
For example, show that log ( 10/1000 ) = log 10 − log 1000.
log (10)
log (1000)
log (10 / 1000) = log (1 / 100)
= log (10^(-2))
= -2
Logarithm law 2
Since a^(1) = a
then, log a (a) = 1
For example
log 2 (2) = 1
log 87 (87) = 1
log 37 (37) + 1 = 38
log 41 (41) * 100 = 100
log 53 (53) * 3 = 3
log x (x) + 2yx = 1 + 2yx
log x (x) / x = 1/x
then, log a (a) = 1
For example
log 2 (2) = 1
log 87 (87) = 1
log 37 (37) + 1 = 38
log 41 (41) * 100 = 100
log 53 (53) * 3 = 3
log x (x) + 2yx = 1 + 2yx
log x (x) / x = 1/x
Logarithm law 1
since a^(0) = 1
then, log a (1) = 0 by definition of logarithm
For example
log 2 (1) = 0
log 150 (1) = 0
then, log a (1) = 0 by definition of logarithm
For example
log 2 (1) = 0
log 150 (1) = 0
Definition of Logarithms
Definition : Loagarithms
If a^(n) = x, then: loga(x) = n, where a >0 ; a =/ 1 and x > 0
The logarithm of a number is the value to which the base must be raised to give the number i.e. the exponent. From the first example of the activity log_2 (16) means the power of 2 that will give 16. As 2^4=16 we see that
log2(16) = 4
If a^(n) = x, then: loga(x) = n, where a >0 ; a =/ 1 and x > 0
The logarithm of a number is the value to which the base must be raised to give the number i.e. the exponent. From the first example of the activity log_2 (16) means the power of 2 that will give 16. As 2^4=16 we see that
log2(16) = 4
피드 구독하기:
글 (Atom)