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Logarithmic properties
| 1. | loga (1) = 0 So when loga (1) = x ⇒ ax = 1 ⇒ x = 0. |
| 2. | loga (a) = 1 So when loga (a) = x ⇒ ax = a ⇒ x = 1. |
| 3. | The domain of the logarithmic function is (0, ∞) Of course, negative numbers have no logarithm, then for eaxample, if you calculate loga (–2), it produces a conflict: loga (–2) = x ⇒ ax = –2 . Because a > 0, the expression ax is always positive for all values of x and ax = –2 gives no solution. |
| 4. | loga(x· y) = loga x + loga y Let loga x = m and loga y = n, then you get am = x y an = y. Therefore |
| 5. | loga (x / y) = loga x – loga y Let loga x = m and loga y = n, then you get am = x and aN = y. Therefore |
| 6. | loga (xn) = n loga x Of course loga (xn) = loga (x · x · x ··· x) = loga x + loga x + ··· + loga x = n loga x |
| 7. | Change of base a in base b Let aN = x ⇒ N = loga x. For the logarithm with base b you get |