Properties of Logarithms Since the exponential and logarithmic functions with base a are inverse functions, the Laws of Exponents give rise to the Laws of Logarithms. Laws of Logarithms: Let a be a positive number, with a ≠ 1. Let A > 0, B > 0, and C be any real numbers. Law Description 1. log a (AB) = log a A + log a B The logarithm of a product of numbers is the sum of the logarithms of the numbers. ⎛⎞A The logarithm of a quotient of numbers is the 2. log a ⎜⎟ = log a A – log a B ⎝⎠B difference of the logarithms of the numbers. C 3. log a (A ) = C log a A The logarithm of a power of a number is the exponent times the logarithm of the number. Example 1: Use the Laws of Logarithms to rewrite the expression in a form with no logarithm of a product, quotient, or power. ⎛⎞5x3 (a) log3 ⎜⎟2 ⎝⎠y (b) ln 5 x2 ()z+1 Solution (a): 3 ⎛⎞5x 32 log333⎜⎟2 =− log() 5xy log Law 2 ⎝⎠y 32 =+log33 5 logxy − log 3 Law 1 =+log33 5 3logxy − 2log 3 Law 3 By: Crystal Hull Example 1 (Continued) Solution (b): 1 ln5 xz22()+= 1 ln( xz() + 1 ) 5 1 2 =+5 ln()xz() 1 Law 3 =++1 ⎡⎤lnxz2 ln() 1 Law 1 5 ⎣⎦ 112 =++55lnxz ln() 1 Distribution 11 =++55()2lnxz ln() 1 Law 3 21 =+55lnxz ln() + 1 Multiplication Example 2: Rewrite the expression as a single logarithm. (a) 3log44 5+− log 10 log4 7 1 2 (b) logx −+ 2logyz2 log( 9 ) Solution (a): 3 3log44 5+−=+− log 10 log 4 7 log 4 5 log 4 10 log 4 7 Law 3 3 =+−log444 125 log 10 log 7 Because 5= 125 =⋅−log44() 125 10 log 7 Law 1 =−log4 () 1250 log4 7 Simplification ⎛⎞1250 = log4 ⎜⎟ Law 2 ⎝⎠7 By: Crystal Hull Example 2 (Continued): Solution (b): 1 1 2222 logxy−+ 2log2 log() 9 z =−+ log xyz log log( 9) Law 3 =−logxy log2 + log() 3 z Simplification =+logx log 3z Law 2 ()y2 () ⎡⎤x = log2 () 3z Law 1 ⎣⎦()y = log3zx Simplification ()y2 Example 3: Evaluate the expression 5 (a) log2 16 (b) log33 189− log 7 Solution (a): 1 5 5 log22 16= log 16 1 = 5 log2 16 Law 3 1 4 4 = 5 log2 2 Because 16= 2 1 = 5 ()4 Property 3 of Logarithms (Section 6.2) 4 = 5 Multiplication Solution (b): 189 log333 189−= log 7 log7 Law 2 = log3 27 Simplification 3 3 = log3 3 Because 27= 3 = 3 Property 3 of Logarithms (Section 6.2) By: Crystal Hull Change of Base: A calculator can be used to approximate the values of common logarithms (base 10) or natural logarithms (base e). However, sometimes we need to use logarithms to other bases. The following rule is used to convert logarithms from one base to another. Change of Base Formula: loga x logb x = loga b Example 4: Use the Change of Base Formula and a calculator to evaluate the logarithm, correct to six decimal places. Use either natural or common logarithms. (a) log6 17 (b) log5 2.33 Solution (a): The Change of Base Formula says loga x logb x = . loga b Thus, if we let the new base a =10 log10 17 log6 17 =≈1.581246 log10 6 Solution (b): Again we will use the Change of Base Formula. This time we will let the new base be a = e. ln 2.33 log 2.33=≈ 0.525568 5 ln 5 By: Crystal Hull Example 5: Simplify: ()log81 12( log2 7) Solution: Using the Change of Base Formula with the new base a =10 : log12 log 7 log 12 = and log 7 = 8 log8 12 log12 Thus, ⎛⎞⎛⎞log12 log 7 ()()log81 12 log2 7 = ⎜⎟⎜⎟ ⎝⎠⎝⎠log8 log12 log 7 = Simplification log8 = log8 7 Change of Base Formula By: Crystal Hull .