Properties of Logarithms Section 4.3 Properties of Logarithms Elementary Algebra Section 4.3 Properties of Logarithms Properties of Logarithms We discuss the common computational properties of logarithms that make them so useful. Included are simple derivations of the product, quotient and power rules. Section 4.5 v5.0.1 10/26/2012

Properties of Logarithms Consider logb x = m and logb y = n By definition bm = x and bn = y xy = (bm)(bn) = bm + n So logb (xy) = logb (bm + n) Product Rule for Logarithms logb xy = logb x + logb y for any positive real numbers b, x, y with b ≠ 1 = m + n WHY? = logb x + logb y Properties of Logarithms: The Product Rule We use the definition of logarithm to identify the log (base b) of two numbers x and y. We call these values m and n, respectively. The definition allows us to write the product xy as a product of two exponential values, bm and bn, to which we then apply the product rule for exponents. This gives us the sum m+n. Using the inverse function property of the logarithm function, or just the definition of logarithm, we find the log of both sides of the equation. This gives the log of the product xy on one side and the sum of the logs on the other side, which is essentially the product rule for logarithms. Note that the transformation from product to sum hinges on the product rule for exponents. The inverse function property for logarithms (or simply the definition of logarithm) allows us to move the sum out of the exponent, where it represents the sum of the logarithms. 10/26/2012 Section 4.5 v5.0.1 Section 4.5 v5.0.1 10/26/2012

Properties of Logarithms Examples 1. log4 (3  7) = log4 3 + log4 7 2. log8 10 + log8 3 = log8 (10  3) = log8 30 3. loga x2 = loga xx = loga x + loga x = 2 loga x 4. 1 + log 2 + log x + log 2x2 5. ln (x – 1) + ln (x + 1) Properties of Logarithms: The Product Rule Examples In Example 1 we simply replace the x and y in the rule by 3 and 7, respectively. The base of 4 can be a challenge for actual computation. We will learn how to convert these forms to forms more calculator-friendly later in this module. In Example 2, we combine a sum of logarithms by using the “reverse” to form of the product rule. In Example 3, we investigate what happens when the argument of the logarithm is raised to a power. The power turns out to be the multiplier of the logarithm function. In Example 4, we again add logarithms, but with variable arguments, using the “reverse” form of the product rule. = 1 + log 4x3 = ln (x – 1)(x + 1) ( ) = ln (x2 – 1) 10/26/2012 Section 4.5 v5.0.1 Section 4.5 v5.0.1 10/26/2012

Properties of Logarithms Again consider logb x = m and logb y = n , for x, y, b positive, b ≠ 1 Thus bm = x and bn = y Quotient Rule for Logarithms for any positive real numbers b, x, y with b ≠ 1 x y logb = bm bn = logb bm – n = m – n WHY? = logb x – logb y Properties of Logarithms: The Quotient Rule This rule follows the same pattern as that for products. However, instead of considering the log of the product of x and y, we consider the quotient x/y. Applying the definition of the logarithm function allows us to write x and y as powers of the common base b. The quotient rule for exponents enables us to exchange the quotient for a difference of exponents, and the inverse function property then allows the conversion of the logarithm of the exponent difference to the difference of logarithms. Notice how intimately the properties of logarithms are related to analogous properties of exponents. logb = logb x – logb y x y 10/26/2012 Section 4.5 v5.0.1 Section 4.5 v5.0.1 10/26/2012

Properties of Logarithms Examples 1. 2. 3. Solve for x : 9 4 ( ) log7 = log7 9 – log7 4 3 16 ( ) log4 = log4 3 – log4 16 = log4 3 – 2 WHY? log (x + 3) log (x + 1) = 2 log (x + 3) log (x + 1) = 2 Properties of Logarithms: The Quotient Rule Examples In Example 1, we use a simple quotient, replacing the x and y in statement of the rule with 9 and 4, respectively. In Example 2, we again apply the quotient rule to produce a difference of logarithms. In this case, the second logarithm is reducible to a constant. In Example 3, we solve a logarithmic equation using first the quotient rule and then a special case of the product rule. We then invoke the 1-1 property of the logarithm function to claim the equality of x +3 and (x + 1)2. Expanding the right side and gathering like terms, we form a polynomial equation, which is easily solvable by factoring and applying the zero-product rule. log (x + 1) = 2 WHY? x + 3 = (x + 1) 2 = x2 + 2x+ 1 = x2 + x – 2 = (x + 2)(x – 1) Solution set: { –2 , 1 } 10/26/2012 Section 4.5 v5.0.1 Section 4.5 v5.0.1 10/26/2012

Properties of Logarithms Consider logb x = m for x, b positive, b ≠ 1 Thus bm = x and xr = bmr So logb xr = logb (brm) Power Rule for Logarithms for any positive real numbers b, x with b ≠ 1 and (bm)r = xr , for any real r = brm = rm WHY? = r logb x Properties of Logarithms: The Power Rule Here again the analogous properties of exponents and the definition of logarithm (the inverse function property) combine to produce the power rule for logarithms. The result is that we can now convert the exponent on the argument of the logarithm to a multiplier of the logarithm without the exponent. This property can be very useful in later work with logarithms. logb xr = r logb x 10/26/2012 Section 4.5 v5.0.1 Section 4.5 v5.0.1 10/26/2012

Properties of Logarithms Examples 1. 2. 3. 3 log5(x + 1) 4. x log 2 5. ln 1 6. ln 0 log352 = 2 log35 loga x4 = 4 loga x = log5(x + 1)3 = log 2x = loge1 = 0 Properties of Logarithms: The Power Rule Examples In Examples 1 and 2 the Power Rule is applied directly to move the exponent to the role of multiplier. In Examples 3 and 4 we apply the Power Rule “in reverse” to move a multiplier to the role of exponent. In Example 5 we simply note a basic rule for all logarithm functions: the log of 1 is always 0, regardless of the base. The definition makes this clear: since the base is never 1, only power of the base that will yield 1 is 0. In Example 6 we ask what power of the (non-zero) base yields 0. Clearly there is no such power. That is, logb0 does not exist for any base b. = ? Question: Is 0 in the domain of any logarithm function ? What does this tell you about ln 0 ? 10/26/2012 Section 4.5 v5.0.1 Section 4.5 v5.0.1 10/26/2012

Properties of Logarithms Property Recognition Rewrite as a logarithm of a single expression : 1. log 4 + log 7 2. log 35 – log 7 3. ln 5e – ln 4. log 5e – log 5. logb x5 – logb x3 + logb x2 6. 7. logb (x  y) in terms of logb x and logb y 1 20e ( ) e 20 ( ) Property Recognition Examples 1 and 2 use the Product Rule and Quotient Rule, respectively, to produce log 28 and log 5 . Examples 3 and 4 use the Quotient Rule and the properties of fractions to produce ln (100e2) and 2, i.e. log 100, respectively. Example 5 uses both the Product Rule and the Quotient Rule to produce x4. For Example 6 there is no logarithm of a single expression for the given expression. The student should note specifically that and in Example 7 there is simply no way to write the expression in terms of log b x and log b y. logb x logb y 10/26/2012 Section 4.5 v5.0.1 Section 4.5 v5.0.1 10/26/2012

Properties of Logarithms More Examples 1. Rewrite in expanded form: log4 (3x + 7) Cannot be written in expanded form ! 2. TRUE or FALSE : log6 36 – log6 6 = log6 30 Rewriting: log6 (36/6) = log6 (6  5) log6 6 = log6 6 + log6 5 0 = log6 5 Since log6 1 = 0 then log6 1 = log6 5 This implies that 1 = 5 ... a CONTRADICTION !! Hence the given statement is FALSE ! More Examples The expression in Example 1 simply cannot be written in any expanded form of logarithms. This form is often confused with the logarithm of a product, which is why it is included here. In Example 2 we note that the left side can be converted to a single logarithm using the quotient rule for logarithms and the right hand side can be expressed as a sum of logs using the product rule. Subtracting log6 6 (or its equivalent of 1) from both sides of the new equation gives us log6 5 = 0. Since log6 1 = 0, and since the logarithm is a 1-1 function, this implies that 5 = 1, a contradiction. The rules of logic tell us that the only way this contradiction can be obtained is to assume something in the problem that is false – namely that the original equation is a true statement. Hence, that statement must be false. This last is an example of indirect reasoning: assume a proposition is true (or false) and show that the assumption leads to a contradiction. This proves the negation of the assumed proposition. 10/26/2012 Section 4.5 v5.0.1 Section 4.5 v5.0.1 10/26/2012

Properties of Logarithms More Examples 3. TRUE or FALSE : log3 (log2 8) log7 49 log8 64 = log3 (3) log7 72 log8 82 = 1 2 log7 7 2 log8 8 = 1 2 (1) = More Examples Example 3 proposes that the given statement is true. By transforming both sides of the equation to a common form we see that the given statement is in fact true. Note that it is not necessary that each side be reduced to 1 but only that the two sides of the equation be reduced to the same value. 1 = 1 So, the given statement is TRUE !! 10/26/2012 Section 4.5 v5.0.1 Section 4.5 v5.0.1 10/26/2012

Properties of Logarithms Bases for Logarithms Properties of Logarithms Conversions Can we use logb x to find loga x ? Let loga x = y By definition ay = x logb (ay) = logb x y logb a = logb x (loga x)(logb a) = logb x Thus applying logb ... applying power rule replacing y Change of Base Rules Conversion of bases is sometimes necessary, even when using the most sophisticated graphing calculators. For example, we might be able to find the logarithm of some x to one base, say b, but we cannot directly compute the logarithm of x to a second base a. By using the change of base rule, we can find the log of x to any base a using the base b that we know. Once again we employ the definition of logarithm and the inverse function property. We convert the logarithm we seek to a power function form and then apply the inverse function property (using a different known base) to solve for the logarithm value we seek, namely y in this case. We see that the logs of x to two different bases, a and b, are proportional with the constant of proportionality or depending on which base we desire to use for x. Please note the last statement in the illustration: do not confuse the quotient of logarithms with the logarithm of a quotient. loga x logb x logb a = logb x loga x loga b = OR logb x logb a NOTE: logb x – logb a ≠ 10/26/2012 Section 4.5 v5.0.1 Section 4.5 v5.0.1 10/26/2012

Properties of Logarithms Bases for Logarithms Properties of Logarithms Conversion Examples Find log3 17 on your calculator Having trouble ? ... if you can Let’s try using a little math first ... log 17 log 3 = 1.23044 0.477121 = log3 17 ≈ 2.5789 OR ln 17 ln 3 = 2.8332 1.0986 = Base Conversion Examples The examples chosen here illustrate that even use of a calculator might not be very helpful without knowledge of the change of base rule. We wish to find the logarithm of a value to an “awkward” base, say 3. In the first case, finding the base-3 log of 17 requires use of a base not generally available on most calculators. However, if we “do the math” first and convert to some other well known base, say base 10, then we can immediately compute the desired logarithm. The common log of both 17 and 3 are easily found and the quotient of these is the desired logarithm of 17 to base 3. Similarly, we can use the natural logarithm, also commonly available on most graphing or scientific calculators. The conversion rule is the same: find the logarithm of the desired value to a convenient base and then divide by the logarithm of awkward base. log3 17 ≈ 2.5789 10/26/2012 Section 4.5 v5.0.1 Section 4.5 v5.0.1 10/26/2012

Properties of Logarithms More Equations Properties of Logarithms Solve 1. Find x to the nearest whole number e.02x = 192 ln(e.02x) = ln(192) (.02x)ln e = ln(192) .02x = 5.2575 x ≈ 262.9 ≈ 263 2. Find x exactly log3 (x + 1)5 = 3 (x + 1)5 = 27 x + 1 = 271/5 x = –1 + 271/5 Solution set: { 263 } Solving Equations Solving and exponential equations usually requires use of the inverse function property. In Example 1 we seek to the value of x in an exponent. To solve for x, we need to move it out of the exponent, which is exactly what applying the inverse logarithm function will do. Because the exponential function is a base-e function, that is the natural exponential function, we use the natural logarithm (to base e). This allows us to solve for x in terms of the natural log of a constant. Rounding to the nearest whole number gives the solution shown. In Example 2 we must solve for x embedded in the logarithm argument (x+1)5 using base 3. We use the inverse function property by applying the base-3 exponential to both sides of the equation. This removes the explicit reference to the logarithm and yields a simple linear equation in x. To find the exact solution, we avoid expression as a decimal fraction and leave the expression with the fifth root of 27 intact. This is often desirable when this result is an intermediate result in a much larger problem, since we can often improve accuracy by putting off use of approximations until the last step of the solution. 3log (x+1) 3 5 = 33 Solution set: { –1 + 271/5 } 10/26/2012 Section 4.5 v5.0.1 Section 4.5 v5.0.1 10/26/2012

Properties of Logarithms More Equations Properties of Logarithms Solve 3. Find x exactly log8 ( 2x + 5) + log8 3 = log8 33 log8 ( 2x + 5) + log8 3 = log8 3 + log8 11 log8 ( 2x + 5) = log8 11 2x + 5 = 11 x = 3 4. Find x exactly log3 2x – log3 (3x + 15) = –2 = log8 (3 ∙ 11) Solution set: { 3 } More Equations In Example 3 we first note that 33 is just the product of 3 and 11. Rewriting the right hand side of the equation we use the product rule for logarithms to separate log8 3 which is then subtracted from both sides of the equation. We then apply the 1-1 property of the logarithm function to yield a simple linear equation in x. This is easily solved, as shown in the illustration. In Example 4 we first use the quotient rule (in “reverse”) and then apply the inverse function property by exponentiating both sides of the equation, yielding a rational function on the left side of the equation. Clearing the fraction and dividing both sides by 3 –2 gives us a simple linear equation which is easily solved. 3x + 15 2x ( ) log3 = –2 3x + 15 2x = 3–2 18x = 3x + 15 x = 1 Solution set: { 1 } 10/26/2012 Section 4.5 v5.0.1 Section 4.5 v5.0.1 10/26/2012

Properties of Logarithms Think about it ! 10/26/2012 Section 4.5 v5.0.1 Section 4.5 v5.0.1 10/26/2012

Properties of Logarithms Section 4.5 Properties of Logarithms Module 4 Section 4.5 Properties of Logarithms Properties of Logarithms We discuss the common computational properties of logarithms that make them so useful. Included are simple derivations of the product, quotient and power rules. Section 4.5 v5.0.1 10/26/2012