 log 5 + log 8 = log _____  log 36 – log 4 = log _____  log 5 = 2 log _____  log 7 = _____ log 7  1.5 log 13 = log _____  The name “logarithm” comes from the words ________ and __________. 5  8 = ÷ 4 = logos arithmos =

EQUATIONS & OTHER BASES AND FUNCTIONS

 Use logarithms with base 10 or other bases to solve exponential equations.

 Common logarithm  Natural logarithm ee  Change of base  Exponential equation  Logarithmic equation

 If x = 10, then y is the base-10 logarithm of x. Similarly, if x = 2, then y is the base-2 logarithm of x. The only difference is the number that is the base. To distinguish among logarithms with different bases, the base is written as a subscript after the abbreviation log. For instance:  The symbol is pronounced “log to the base 2 of 8”. The symbol is, of course, equivalent to log 100, as defined in the previous sections. Note that in all cases a logarithm is an exponent.

Logarithm with any Base Algebraically: if and only if, where b > 0 b ≠ 1 and x > 0 Verbally means that y is the exponent of b that gives x as the answer

 The way you pronounce the symbol for logarithm gives you a way to remember the definition. Examples 1 and 2 show you how to do this. Write in exponential form. Solution: Think this:  “log … is read “log to the base 5…,” so 5 is the base  A logarithm is an exponent. Because the log equals a, a must be the exponent.  The “answer” I get for 5 is the argument of the logarithm, c. Write only this:

Write in logarithmic form. Solution:  Two bases of logarithms are used frequently enough to have their own key on most calculators. One is base-10 logarithms, or common logarithms. The other is base-e logarithms, called natural logarithms, where e = …, a naturally occurring number (like π ) that you will find advantageous later in your mathematical studies. The symbol ln x (pronounced “el en of x”) is used for the natural logarithms: ln x =

Common Logarithm and Natural Logarithm Common: The symbol log x means Natural: The symbol ln x means, where e is a constant equal to …

 To find the value of a base-e logarithm, just press the ln key. For instance: ln 30 = …. To show what this answer means, raise e to the …power. Use the e key. Do not round the … Example 3 shows you how to find a logarithm with a base that is not built into your calculator.

 Find. Check your answer by an appropriate numerical method. Use the definition of logarithm. Solution: Let 5 x = 17 log 10 5 x = log Take log 10 of both sides. x log 10 5 = log Use the log of a power property to “peel off” the exponent. x = = … Divide both sides by the coefficient of x. log 5 17 = … Substitute for x. Check: … = 17 Do not round the ….

 In example 3, note that the base-5 logarithm of a number is directly proportional to the base-10 logarithm of that number. The conclusion of the example can be written this way:  To find the base-5 logarithm of any number, simply multiply its base-10 logarithm by …(that is divide by log 5 17 = · log = … log  This proportional relationship is called the change-of- base property. From the result of example 3, you can write:

Property: The Change-of-Base Property of Logarithms or Notice that the logarithm with the desired base is by itself on the left side of the equation and that the two logarithms on the right side have the same base,presumably one available on your calculator. The box shows this property for bases a and b and argument x.

 Find ln 29 using the change-of-base property with base-10 logarithms. Check your answer directly by pressing ln 29 on your calculator. Press log (29)/log (e). Solution: ln 29 = Directly: ln 29 = … Which agrees with the value you got using the change-of-base property

The properties of base-10 logarithms presented in the previous section are generalized here for any base. The Logarithm of a Power log b x y = y log b x Verbally: The logarithm of a power equals the product of the exponent and the logarithm of the base. Log of a Product. log b (xy) = log b x + log b y Verbally: The logarithm of a product equals the sum of the logarithms of the factors Log of a Quotient log b = log b x − log b y Verbally: the logarithm of a quotient equals logarithm of numerator minus logarithm of denominator

 Logarithms provide a way to solve an equation with a variable in the exponent or to solve an equation that already contains logarithms. Examples 5-8 show how you can do this. Example 5: Solve the exponential equation = 983 algebraically, using logarithms. Solution: = 983 log = log 983 Take the base-10 logarithm of both sides. 3x log 7 = log 983 Use the logarithm of a power property. Divide both sides by the coefficient of x. x = …

 Solve the equation: Apply the logarithm of a product property Solution: log 2 (x − 1) + log 2 (x − 3) = 3 log 2 (x − 1)(x − 3) = = (x − 1)(x − 3) Use the definition of a logarithm. 8 = x 2 −4x + 3 Expand the product x 2 − 4x − 5 = 0 Reduce one side to zero. Use the symmetric property of equality. (x − 5)(x + 1) = 0 Solve by factoring. x 1 = 5 or x 2 = −1 Solutions of the quadratic equation.

 You need to be cautious here because the solutions in the final step are the solutions of the quadratic equation, and you must make sure they are also the solutions of the original logarithmic equation. Check by substituting your solutions into the original equation. If x 2 = -1 thenIf x 1 = 5 then log 2 (5 − 1) + log 2 (5 – 3) = log log = 3 log 2 (-1 – 1) + log 2 (-1 – 3) = log 2 (-2) + log 2 (-4) which is undefined. x 1 = 5 is a solution, but x 2 = -1 is not

 Solve the equation and check your solutions. Apply the properties of exponents. Solution: You can realize that this is a quadratic equation in the variable. Using the quadratic formula, you get: or

 You now have to solve these two equations. Check: x 1 = ln 2 = x 2 = 0 e 2 ln 2 − 3e ln = (e ln 2 ) 2 − 3e ln = 2 2 − 3(2) + 2 = 0 Both solutions are correct.

 Solve the logarithmic equation ln (x + 3) + ln (x + 5) = 0 and check your solution(s). Use the logarithm of a product property Solution: By quadratic formula x + 8x + 15 = 1 ln (x + 3) + ln (x + 5) = 0 ln [(x + 3)(x + 5)]= 0 (x + 3)(x + 5)= e = 1 x + 8x + 14 = 0 x = …or x = …

Check: ln ( …+ 3) + ln ( …+ 5) = … … = 0 x = … ln ( … + 3)+ ln ( … + 5) = ln (0.4142…) + ln (2.4142…) which checks x = … ln ( …) + ln ( …) which is undefined The only valid solution is x = …

 Textbook pg. 326 #2-48 Every other even