Basic Introduction To Logarithm

A logarithm is used to solve an exponential equation, where the unknown is the exponent. Think of using multiplication to solve a division problem. The logarithm of 1000 to base 10 is 3, because 10 to the power 3 is 1000: 1000 = 10 × 10 × 10 = 10³.The logarithm to base 10 (b = 10) is called the common logarithm and has many applications in science and engineering. y = b^x can be rewritten as log_b y = x So, when you compute log 3, you are actually solving for x in 10^x = 3. For example, to solve for x in 10^x = 25, rewrite using log. 10^x = 25 b is 10, y is 25, and x is our unknown exponent log₁₀ 25 = x      the log button on your calculator is actually log10 , so you can just push log 25 and that is your answer log₁₀ 25 = 1.398, so x = 1.398 To check your answer, put 1.398 back into the original problem: 10^1.398 ≈ 25 Some properties of logarithms are given below: If a^m = x, then m = loga(x). Properties : a. log_b(b) = 1 b. log_b(1) = 0 c. log_b(mn) = log_b(m) + log_b(n) d. log_b(^m/_n) = log_b(m) – log_b(n) e. log_b(m) = 1/log_m(b) f. log_b(mⁿ) = n · log_b(m) g. log_b(m)= log_a(m)/log_a(m) A logarithm for base 1 does not exist. Remember that, when base is not mentioned, it is taken as 10. In less formal terms, the log rules might be expressed as: 1) Multiplication inside the log can be turned into addition outside the log, and vice versa. 2) Division inside the log can be turned into subtraction outside the log, and vice versa. 3) An exponent on everything inside a log can be moved out front as a multiplier, and vice versa. Warning: Just as when you’re dealing with exponents, the above rules work only if the bases are the same. For instance, the expression “log_d(m) + log_b(n)” cannot be simplified, because the bases (the “d” and the “b”) are not the same, just as x² × y³ cannot be simplified (because the bases x and y are not the same).  Expanding logarithms: Log rules can be used to simplify expressions, to “expand” expressions, or to solve for values.   Expand log₄( ¹⁶/_x ). I have division inside the log, which can be split apart as subtraction outside the log, so log₄( ¹⁶/_x ) = log₄(16) – log₄(x) The first term on the right-hand side of the above equation can be simplified to an exact value, by applying the basic definition of what a logarithm is: log₄(16) = 2 Then the original expression expands fully as: log₄( ¹⁶/_x ) = 2 – log₄(x) Always remember to take the time to check to see if any of the terms in your expansion (such as the log₄(16)above) can be simplified.