Also, read |
Log Rules |
Logarithm:
Earlier we have studied exponents which covers every part of mathematics. Now it’s contained to know that logarithm is the inverse function of exponentiation. In this article, we will read about the log formula. Logarithm is denoted as the repeated multiplication of the same factor. As any number as 81 can be factorized of same factors 3 as 3 x 3 x 3 x 3 which is 4 times multiplication of 3. Now we can say that log of 81 for factor (base) 3 is 4.
i.e. log3(81) = 4 here, 3 is base.
Points:
- We can identify the logarithmic value of positive numbers only.
- log(0) = undefined
- The logarithmic value of negative numbers is imaginary. There is no real value of log function for negative numbers.
Log Formula:
S. no. | Log Formula | Log Rules |
1. | loga(m.n) = loga m + loga n | Product rule |
2. | loga( \(\frac{m}{n}\) ) = loga m – loga n | Quotient rule |
3. | loga a = 1 | Log with same base |
4. | loga 1 = 0 | Log of 1 |
5. | loga (xn) = n loga x | Log of power |
6. | loga x = \(\frac{1}{log_{x}(a)}\) | Log of reciprocal |
7. | Logax = \(\frac{log_{b}(x)}{log_{b}(a)}\) = \(\frac{ log\:x}{log\:a}\) | Log of distribution |
8. | loga(ak) = k | Different power with same base |
9. | alogb(k) = k | Component rule |
10. | logak x = \(\frac{1}{x}\) loga x, for k ≠ 0 | Base exponent rule |
In all these rules a>1, m, n, k be any real number but m, n must be positive. Then log formula will be valid.
Terms related to log formula:
- Common logarithm: The log which has base of 10 is known as common log.
Example: log(24) = log10(24) - Natural logarithm: The log which has base of e (i.e. Euler,s number has a value of 2.71) is known as natural log. It is denoted by “ln”.
Example: ln(24) = loge(24) - Negative logarithms: It has a simple meaning that how many times we divide a number.
Example: log8(0.125) = -1, as 1/8 = 0.125
Log formula Examples:
- Solve the sum of log10(25) + log10(4).
Solution:
we know that loga(m.n) = loga m + loga n
so, log10(25) + log10(4)
= log10(100)
= log10(10)2
= 2 log10(10) = 2
- Solve log5500 −2 log52.
Solution:
we can write log5500 −2 log52 as log5500 −2 log52
= log5500 − log5 (2)2
= log5500 − log5 (4)
As we know that loga m – loga n = loga( \(\frac{m}{n}\) )
= log5( \(\frac{500}{4}\) )
thus, log5(125)
= log5(5)3 = 3
More from Calculus | |
Relation and Functions | Limits Formula |
Continuity Rules | Differentiability Rules |
Derivative Formula | Integral Formula |
Inverse Trigonometric function Formulas |
Application of Integrals |
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